GRE Math Essentials - A most comprehensive handout!!

­TRIANGLES





A closed figure enclosed by 3 sides is called a Triangle. ABC is a triangle. The sides AB, BC, AC are respectively denoted by c, a, b. Please carefully note the capital and small letters.

In any triangle ABC

(1) \(A = \frac{1}{2} a \times h = \frac{1}{2} base\)  \(\times\) perpendicular to base from opposite vertex
(2) \(A = \sqrt{s(s –a)(s – b)(s – c)}\) , \(s = \frac{(a + b + c)}{2} =\) semi–perimeter
(3) \(P = (a + b + c) = 2 s\)


PROPERTIES :
 1. Sum of the three interior angles is 180°
 2. When one side is extended in any direction, an angle is formed with another side. This is called the exterior angle.
 3. There are six exterior angles of a triangle.
 4. Interior angle + corresponding exterior angle = 180°.
 5. An exterior angle = Sum of the other two interior angles not adjacent to it
 6. Sum of any two sides is greater than the third side.
 7. Difference of any two sides is less than the third side.
 8. Side opposite to the greatest angle will be the greatest and vice versa.
 9. A triangle must have at least two acute angles.
10. Triangles on equal bases and between the same parallels have equal areas.
11. If a, b, c denote the sides of a triangle then

(i) if \(c^2 < a^2 + b^2\), Triangle is acute angled
(ii) if \(c^2 = a^2 + b^2\), Triangle is right angled
(iii) if \(c^2 > a^2 + b^2\), Triangle is obtuse angled


Right Angled Triangle



A triangle whose one angle is 90° is called a right (angled) Triangle. In the figure, b is the hypotenuse, and a & c the legs, called base and height resp.

(1) \(h^2 = m n\)
(2) \(AC^2 = AB^2 + BC^2\) (Pythagoras theorem)
(3) \(h = \frac{ac}{b}\)
(4) \(Area = \frac{ac }{ 2}\)

NOTE : You should remember some of the Pythagorean triplets (e.g. 3,4,5 because \(5^2 = 3^2 + 4^2\)). Some others are (5, 12, 13), (7, 24, 25) etc.




Isosceles Right Triangle




A right triangle whose two legs are equal is an isosceles right triangle. The ratio of sides is \(1 : 1 : \sqrt{2}\)


30, 60, 90 triangle



This is a special case of a right triangle whose angles are 30°, 60°, 90°.

In this triangle side opposite to angle \(30° = \frac{HYP}{2}\), Side opposite to Angle \(60° =\) \(\frac{\sqrt{3}}{2} \times HYP\). The ratio of sides is \(1 : \sqrt{3} : 2\)



Equilateral Triangle



A triangle whose all sides are equal is called an equilateral triangle. If a be the side of an equilateral triangle, then

1. Area \(= \frac{\sqrt{3}}{4} a^2\)

Altitude \(= h=\frac{\sqrt{3}}{2}a\)

2. Given the perimeter, equilateral triangle has the maximum area.

3. Of all the triangles that can be inscribed in a circle, the equilateral triangle has the greatest area.

4. Area of Outer Circle is 4 times the area of Inner Circle

5. The triangle formed by joining the mid-points of the sides will be half in perimeter and one-fourth in area

6. For similar figures and solids, \(\frac{A_1}{A_2} = \left( \begin{array}{cc}    \frac{\ell_1}{\ell_2 }            \end{array} \right)^2\) and for similar solids \(\frac{V_1}{V_2} = \left( \begin{array}{cc}    \frac{\ell_1}{\ell_2 }            \end{array} \right)^3\)

7. A triangle with 2 sides (and two angles) equal is called isosceles.



GENERAL THEOREMS ON SIMILARITY

Proportionality Theorem



Intercepts made by two transversal lines (cutting lines) on three or more parallel lines are proportional. In the figure, lines \(X_1, Y_1,\) & \(X_2Y_2\) are transversals cutting the three parallel lines AB,CD, EF. Then AC, CE, BD, DF are intercepts

Also, \(\frac{AC}{BD} = \frac{CE}{DF}\)


Midpoint Theorem

A triangle, the line joining the mid points of two sides is parallel to the third side and half of it.


Basic Proportionality Theorem




A line parallel to any one side of a triangle divides the other two sides proportionally. If DE is parallel to BC, then

(a) \(\frac{AD}{BD} = \frac{AE}{EC}\),
(b) \(\frac{AB}{AD} =\frac{ AC}{AE}\),
(c) \(\frac{AD}{DE} = \frac{AB}{BC}\) and so on.



PROPERTIES OF SIMILAR TRIANGLES

(1) Ratio of areas = Ratio of squares of corresponding sides
(2) RIGHT TRIANGLE :



ABC is a Right Triangle with A as the Right angle. AD is perpendicular to BC then

(a) Triangle \(ABD ~ Triangle CBA\) & \(BA^2 = BC \times BD\)
(b) Triangle \(ACD ~ Triangle BCA\) & \(CA^2 = CB \times CD\)
(c) Triangle \(ABD ~ Triangle CAD\) & \(DA^2 = DB \times DC\)



Some More Important Points:

1. If with a given perimeter, different figures are formed like equilateral triangle, square, regular hexagon, regular octagon ...... and eventually a circle (a regular polygon of infinite sides), then the triangle will have the minimum area and circle will have the maximum area.
2. If different triangles are inscribed in a circle, then the equilateral triangle will have the maximum area.
3. If the perimeter of a triangle is fixed, then the equilateral triangle will have the maximum area.
4. If the sum of two sides of a triangle is constant, then the isosceles right angled triangle will have the maximum area.

­QUADRILATERALS


A closed figure (plane) bounded by four sides is called a quadrilateral. In any quadrilateral :

\(A = \frac{1}{2} \times\) one diagonal \(\times\) (sum of perpendiculars to it from opposite vertices).

(a) Sum of the four interior angles = 360°
(b) If a quadrilateral can be inscribed in a circle, it is called a cyclic quadrilateral. Here opposite angles are supplementary. If one side is produced, then the exterior angle = Remote interior angle.


Rectangle



A quadrilateral whose opposite sides are equal and each internal angle equal to 90°, is called a rectangle.

\(l = length\) 
\(b = breadth\) 
\(A = l \times b\) 
\(P = 2 (l + b)\)

Diagonal \(= \sqrt{(l^2 + b^2)}\), diagonals are equal and bisect each other.

(a) Opposite sides equal, each angle = 90°
(b) Diagonals bisect each other (not at 90°).
(c) Of all rectangles of given perimeter, a square has max. area
(d) When inscribed in a circle, it will have maximum area when it’s a square.
(e) Figure formed by joining the midpoints of a rectangle is a rhombus.
(f) The biggest circle that can be inscribed in a rectangle will have the diameter equal to the breadth of the rectangle.



(g) When a rectangle is inscribed in a circle, the diameter of the circle is equal to the diagonal of the rectangle.



Parallelogram




A quadrilateral in which opposite sides are equal and parallel is called a parallelogram.

\(A = bh\) 
\(P = 2 (a + b)\), Diagonals bisect each other.

(a) Opposite angles are equal
(b) Diagonals bisect each other (not at 90°)
(c) Sum of any two adjacent angles = 180°
(d) Diagonals are unequal in lengths and do not bisect angles at vertices.



Square



A quadrilateral whose all sides are equal and all angles 90° is called a square

\(Area=a^2 \)

\(Diagonal=a\sqrt{ 2} \)

\(Area=\frac{Diagonal^2}{2}\)

\(P=4 \times side\)


(a) Diagonals bisect each other at 90° and are equal
(b) When inscribed in a circle, diagonal = diameter of circle
(c) When circumscribed about a circle, Side of square = Diameter of circle.
(d) The figure formed by joining the mid-points of the sides of a square is also a square. In this case the side will become \(\frac{1}{\sqrt{2}}\) times the side of the original square, perimeter will become \(\frac{1}{\sqrt{2}}\)  times the perimeter of the original square and area will become \(\frac{1}{2}\) times the area of the original square.
(e) If a square (biggest possible) is inscribed in a circle of radius r, then Diameter of circle = Diagonal of Square. If the area of the circle is A then \(A = \pi r^2=\frac{\pi D^2}{4} or \frac{D^2}{2} =\frac{2A}{\pi}\) or Area of Square \(= \frac{2A}{\pi} or 2r^2\)
(f) If a circle is inscribed in a square of side a, then Side of square = Diameter of circle. If the area of the square is S, then the area of circle \(= \frac{\pi S}{4} =\frac{ \pi a^2}{4}\)



Rhombus

A parallelogram whose all sides are equal is called a rhombus.




\(A=\frac{1}{2} \times D_1 \times D_2 (  D_1, D_2 are diagonals)\)

\(P=4a\)

Diagonals are at right angles 

\(D_1^2+D_2^2=4a^2\)

(a) Opposite angles are equal
(b) Diagonals bisect each other at 90°.
(c) Sum of any two adjacent angles = 180°.
(d) Diagonals are unequal.



Trapezium



A quadrilateral in which one pair of opposite sides is parallel is a trapezium. \(A =\frac{1}{2} \times (a + b) \times h\)



CYCLIC QUADRILATERAL




1. The four vertices lie on a circle.
2. Opposite angles are supplementary.
3. If any one side is produced, Exterior angle = Remote interior angle
 

­Regular Polygons


A many sided closed figure is called a polygon. If all the sides of a polygon are equal, it is called a regular polygon.




1. Interior Angle + Exterior angle = 180°
2. \(P = n \times a\) (n = number of sides, a = side)
3. Sum of exterior angles \(= 360°\)
4. Sum of Interior angles \(= (2n – 4) 90°\)
5. Each Interior angle \(= [(\frac{2n – 4)}{n}] \times 90°\)
6. Each exterior angle \(= \frac{360}{n}\)
7.\( A = \frac{1}{2} P\)
\( r = \frac{1}{2 }\) n a r
P = perimeter (r = ⊥ from centre to any one side = radius of encircle)


Regular Hexagon

1. Each interior angle = 120°
2. Sum of Interior angles = 720°
3. Each exterior angle = 60°
4. Area \(= (\frac{3\sqrt{3}}{2}) \times a^2 (a = side)\)
5. P = 6a


Circle

A circle is the path traversed by a point which moves in such a way that its distance from a fixed point always remains constant.





C is called centre and R the radius of the circle. Circumference \(= 2 \pi R = \pi D\)

1. Area \(= A = \pi R^2 = \frac{\pi D^2}{4} (D = Diameter = 2R) \pi = \frac{22}{7} or 3.14\)
2. Length of arc of a circle is given by \(\ell (AB) = (\frac{θ°}{360°}) \times 2 \pi R\)
3. Area of Sector \(ABC = (\frac{θ}{360}) \times \pi R^2 = \frac{\frac{1}{2}}{(AB) \times R}\)
4. Distance travelled by a wheel in n revolutions \(= n\)\( \times\) circumference
5. Angle at the centre made by an arc = twice the angle made by the arc at any point on the remaining part of the circumference.



We have \(\angle APB =\frac{1}{2} \angle AOB = 30° = \angle AQB\)
6. Angle in a semicircle is a right angle.
 

­I have posted a video on YouTube to discuss about Circles : Basics and Properties



Attached pdf of this Article as SPOILER at the top! Happy learning!

Following is Covered in the Video

Theory

¤ What is a Circle?
¤ Circle Geometry Definitions
¤ Circle : Area and Circumference
¤ Semicircle : Area and Circumference
¤ Arc of a Circle
¤ Sector of Circle
¤ Properties of Circles

What is a Circle?

A Circle is a 2D figure which is formed by joining all the points in a 2D plane which are at a fixed distance (i.e radius) from a single point. (i.e center of the circle)



Circle Geometry Definitions

¤ Radius – A line segment joining the center of the circle to any point on the circle. (Ex: OA)
¤ Chord– A line segment whose two end points lie on the circle (Ex: BC)
¤ Diameter– A chord which passed through the center. (Ex: DE)
     ( Diameter = 2* Radius )



¤ Secant– A line which cuts the circle at two points. (Ex: line s)
¤ Tangent– A line which touches circle at only one point. (Ex: line t)


Circle : Area and Circumference

Area of a Circle with radius r, A = ∏ \(r^2\)



Circumference of a Circle with radius r, C = 2 ∏ r

Central Angle = 360˚


Semicircle : Area and Circumference

Area of a Semicircle with radius r, A = \(\frac{∏r^2}{2}\)



Circumference of a Semicircle with radius r, C =  ∏ r + 2r

Central Angle = 180˚


Arc of a Circle

Arc of a circle is a part of the Circumference of the circle.



Length of Arc AB, which subtends angle Θ at the center,

    AB = \(\frac{𝛩}{360˚}\)∗ 2 ∏ r


Sector of Circle

Sector of a circle is a part of the circle made by the arc of the circle and the two radii connecting the arc to the center of the circle.



Area of sector OACB, which subtends angle Θ at the center,
                        Area of OACB =\( \frac{𝛩}{360˚}∗  ∏ r^2\)

Circumference of Sector OACB is given by
                        Circumference of OACB = \(\frac{𝛩}{360˚}\)∗ 2 ∏ r + 2r


Properties of Circles

PROP 1: A Chord subtends same angle at any point on the circle.



PROP 2: Angle subtended by the chord at the center is twice the angle subtended by the chord at any other point on the circle.



PROP 3: Diameter subtends 90˚ at any point on the circle



PROP 4: From an external point there are only two tangents which can be drawn to a circle and the length of these tangents is equal.



PROP 5: A tangent always makes 90˚with the line joining the point of tangency (point of intersection of the tangent with the circle) to the center of the circle.



PROP 6: Cyclic Quadrilateral
A quadrilateral whose all 4 vertices lie on the circumference of the circle is called a Cyclic Quadrilateral.



Sum of all the angles of Cyclic Quadrilateral = 360 ˚
        ∠ A + ∠ B + ∠ C + ∠ D = 360 ˚

Sum of diagonally opposite angles = 180 ˚
        ∠ A + ∠ C = 180 ˚
        ∠ B + ∠ D = 180 ˚

PROP 7: Perpendicular drawn from the center of the circle to a chord bisects the chord.



PROP 8: Equal chords are equidistant from the center. Or
                        
Chords which are equidistant from the center are equal.



PROP 9: Equal chords subtend same angle at the center of the circle. Or
                        
Chords which subtend same angle at the center of the circle are equal.



Hope it Helps!­

­SOLIDS

Cube

A six–faced solid figure with all faces equal and adjacent faces mutually perpendicular is a cube.



If “a” be the edge of a cube,
1. The longest diagonal \(= a\sqrt{3}\) The face diagonal \(= a\sqrt{2}\)
2. Volume \(= a^3\)
3. Total surface area \(= 6 a^2\)


CUBOID or RECTANGULAR BOX



If a,b,c are the edges of a box,
1. The longest diagonal \(= \sqrt{(a^2 + b^2 + c^2)}\)
2. Surface area \(= 2 (ab + bc + ac)\)
3. Volume \(= abc\)


RIGHT CIRCULAR CYLINDER

If r is the radius of base and h is the height, then

[img]https://gre.myprepclub.com/forum/download/file.php?mode=view&id=14856&sid=39c06ed135b8f48a3a154da208f02f27[/img

1. Volume = \(\pi r^2 h\)
2. Curved surface area \(= 2 \pi rh\)
3. Total surface area \(= 2 \pi r (r + h)\)
4. If a rectangle of length L and breadth B is rotated about its length to form a cylinder, then \(L = 2 \pi R\) and \(B = h\).
5. If a rectangle of length L and breadth B is rotated about its breadth to form a cylinder, then \(B = 2 \pi R\) and \(L = h\)


RIGHT CIRCULAR CONE



R = radius of base 
H = Height 
L = slant height \(= \sqrt{(H^2 + R^2)}\)

1. Volume \(= \frac{1}{3} \times (\pi R^2H)\)
2. Curved surface Area \(= \pi R L\)
3. Total Surface Area \(= \pi R (R + L)\)


SPHERE



R = Radius

1. Volume \(= \frac{4}{3} \times \pi R^3\)
2. Surface Area \(= 4 \pi R^2\)
 

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1. DISTANCE BETWEEN TWO POINTS : If there are two points \(A (X_1, Y_1)\) and \(B (X_2, Y_2)\) on the XY plane, the distance between them is given by \(AB = d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
Distance of the point \(P(x, y)\) from the origin O is \(OP = \sqrt{(x^2+ y^2)}\)

2. The coordinates of \(P(X,Y)\) such that P divides the line joining A (X, Y1) and B (X2, Y2) internally I the ratio m : n will be : \(X=\frac{mX_2+nY_1}{m+n}, Y=\frac{mY_2+nY_1}{m+n}\\
\)

3. If \(P (X,Y)\) is the midpoint between A & B then \(X = \frac{(X_1+ X_2) }{2} , Y = \frac{(Y_1+ Y_2) }{2}\)


EQUATION OF A CURVE :

An equation in two variables X and Y (with or without a constant term) is called the equation of a particular curve if the graph of that equation plotted on the XY cartesian plane gives that particular curve. e.g. \(X^2 + Y^2 = 36\). If we plot this curve, by taking different values of Y (and thereby different values of X), we get a circle. Hence \(X^2 + Y^2 = 36\) is the equation of a circle.

STRAIGHT LINE :

An equation of the form AX + BY + C = 0 is called the general equation of a straight line, where X and Y are variables and A, B, C are constants. Any point lying on this line will satisfy the equation of the line. i.e. the coordinates of the point when substituted by X & Y resp. in the above equation will make the LHS vanish.


GENERAL CONCEPTS

If AB is a straight line on the XY plane, then the ratio of y-intercept to x-intercept (with signs) is called its slope and is denoted by ‘m’. The lengths OP and OQ are respectively called the intercepts on X and Y axes, made by the line.



The same straight line can be represented by many different forms of the same equation.

1. If ‘m’ is the slope of the line and ‘c’ the intercept made by the line on Y axis, the equation is \(Y = mX + c\)

2. If the slope of the line is m and it passes through \((X_1, Y_1)\), the equation is \((Y – Y_1) = m (X – X_1)\)

3. If the line passes through two points \((X_1, Y_1)\) and \((X_2, Y_2)\), the equation is

\(Y-Y_1=\frac{Y_2-Y_1}{X_2-X_1} (X-X_1)\)

Hence the slope of the line through \((X_1, Y_1), (X_2, Y_2)\) is given by \(m = \frac{(Y_2 – Y_1) }{(X_2 – X_1)}\) or \(\frac{(Y_1 – Y_2)}{(X_1 – X_2)}\)


4. If the X intercept of the line is a and Y intercept is b, the equation is : \(\frac{ X}{a }+ \frac{Y}{b} = 1\\
\)

5. General form : \(A_X + B_Y + C = 0\) 

In this slope \(= m = –\frac{A}{B}\) 

X intercept \(= –\frac{C}{A }\)

Y intecept \(= –\frac{C}{B}\)

(a) If two lines are parallel then their slopes are equal \((m_1= m_2)\). If two lines do not intersect, they are parallel.
(b) If two lines are perpendicular to each other, the product of their slopes is –1. \((m_1 m_2 = –1)\).
(c) \(a_1X + b_1Y + c_1 = 0 \) and \(a_2X + b_2Y + c_2 = 0\) will represent the same straight line if \(\frac{a_1}{a_2} = \frac{b_1}{b_2 }= \frac{c_1}{c_2}\) . In this case, the lines are coincident and theoretically intersect at infinite points.

6. The point of intersection of two lines (X,Y) is obtained by simultaneously solving both the equations.

7. The equation of a line parallel to a given line \(AX + BY + C = 0\), will be \(AX + BY + K = 0\) where K is a constant which can be found by any additional given condition.

8. The equation of a line perpendicular to a given line \(AX + BY + C = 0\) will be \(BX – AY + K = 0\), where K is a constant which can be found by additional given conditions.

9. The length of perpendicular (p) from \((X_1, Y_1)\) on the line \(AX + BY + C = 0\) is :




10. Equation of a line parallel to X axis is Y = b (b is a constant)

11. Equation of a line parallel to Y axis is X = a (a is a constant)

12. Equation of X and Y axes are Y = 0 and X = 0 respectively

13. Any point on the X-axis can be taken as (a,0)

14. Any point on the Y-axis can be taken as (0,b)

15. In order to find the X–intercept of a line, put Y = 0 in the equation of the line and find X

16. In order to find the Y-intercept of a line, put X = 0 and find Y

17. The image of the point (a, b) in the x-axis is (a, -b)

18. The image of the point (a, b) in the y-axis is (-a, b)

19. The image of the point (a, b) in the line y = x is (b, a)

20. To plot a line, first put y = 0, find the point on the x-axis; then put x = 0, fins the point on the y-axis. Join the two points to get the desired graph.

Note :
The equation of a circle with centre (h, k) and radius r is \((x – h)^2+ (y – k)^2= r^2\).
The equation of a circle with centre (0, 0) and radius r is \(x^2+ y^2= r^2\).
The equation \(y = Ax^2+ Bx + C\) is the equation of the quadratic graph which is a parabola with an axis parallel to the y-axis. If \(A > 0\), the parabola opens upwards. If \(A < 0\), the parabola opens downwards.
If \(B^2> 4AC\), the parabola cuts the x-axis at 2 different points.
If \(B^2= 4AC\), the parabola touches the x-axis at one point (the two points become co-incident).
If \(B^2< 4AC\), the parabola does not cut the x-axis at all.

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­1. Definition. 

A function of a real variable x with domain D is a rule that assigns a unique real number to each number x in D. Functions are often given letter names such as f, g, F, or φ. We often call x the independent variable or the argument of f. If g is a function and x is a number in D, then g(x) denotes the number that g assigns to x. We sometimes make the idea that F has an argument (we substitute a number for the variable in F) explicit by writing F(·). In the case of two variables we sometimes use y = f(x) for the value of f evaluated at the number x. Note the difference between φ and φ(x)

2. The domain of a function. 

The domain is the set of all values that can be substituted for x in the function f(·). If a function f is defined using an algebraic formula, we normally adopt the convention that the domain consists of all values of the independent variable for which the function gives a meaningful value (unless the domain is explicitly mentioned).

3. The range of a function. 

Let g be a function with domain D. The set of all values g(x) that the function assumes is called the range of g. To show that a number, say a, is in the range of a function f, we must find a number x such that f(x) = a. Here are some example functions for which to find the domain and the range.

1: \(f(x) = x, 0 ≤ x ≤ 60\)

2: \(g(x) = \frac{x^2}{20}, 0 ≤ x ≤ 60\)

3: \(h(x) = \frac{12}{x^2}\)

4: \(φ(x) = 3x^2\)

5: \(x = −\frac{1}{2} y^2, y ≥ 0\)

6: \(f(x) = \frac{3x}{x^2 − 4}\)

7: \(g(x) = \sqrt{4 − 3x}\)


4. The graph of a function. 

When the rule that defines a function f is given by an equation in y and x, the graph of f is the graph of the equation, that is the set of points (x, y) in the xy-plane that satisfies the equation. Another way to say this is that the graph of the function g is the set of all point (x, g(x)), where x belongs to the domain of g.

5. The vertical line test. 

As set of points in the xy-plane is the graph of a function if and only if a vertical line intersects the graph in at most one point.



Compound Functions


The composition of two functions g and f is the new function we get by performing f first, and then performing g. For example, if we let f be the function given by \(f(x) = x^2\) and let g be the function given by \(g(x) = x + 3\), then the composition of g with f is called gf and is worked out as 

\(gf(x) = g(f(x))\)

So we write down what f(x) is first, and then we apply g to the whole of f(x). In this case, if we apply g to something we add 3 to it. So if we apply g to \(x^2\), we add three to \(x^2\). So we obtain

\(gf(x) = g(f(x)) = g(x^2) = x^2 + 3 \)

The order in which we compose functions makes a big difference to the result. You can see this if we change the order of the functions in the first example. We have taken \(f(x) = x^2\) and \(g(x) = x + 3\). Then fg(x) is given by taking g(x), which is x + 3, and applying f to all of it. This gives us

\(fg(x) = f(x + 3) = (x + 3)^2 = x^2 + 6x + 9 \)

You can see that this is not the same as gf(x), because 

\(gf(x) = x^2 + 3\)

and this does not in general equal \(x^2 + 6x + 9\)


source: https://www.mcckc.edu/
https://www.mathcentre.ac.uk/

­

Permutation Vs. Combination
Factor	PERMUTATION	COMBINATION
Meaning	Permutation can be defined as the different ways how we can arrange a set of objects in a sequence.	Combination means the different ways of choosing variables or items from a set of objects not focusing on the order of the same.
Order	The focus is on the order of the positioning of the variables or items	Order is irrelevant in combinatorics.
Denotes	The arrangement of variables	The selection of items in no particular order
What does it show us?	Set of elements in a logical sequence	Sets of items without any order
Answers	Permutation tells us how many groups can be created from a set of objects.	Combinatorics tells us how many different groups can be selected from a larger group of items.

Example : Suppose, there are four quantities A,B,C,D. The different orders of arrangements of these four quantities by taking three at a time, are :

ABC, ACB, BAC, BCA, CAB, CBA, ... (1)
ABD, ADB, BAD, BDA, DAB, DBA, ... (2)
ACD, ADC, CAD, CDA, DAC, DCA, ... (3)
BCD, BDC, CDB, CBD, DBC, DCB. ... (4)

Thus, each of the 24 arrangements, of the four quantities A,B,C,D by taking three at a time, are each called a permutation. Hence, it is clear that the number of permutations of four things taken three at a time is 24. Hence, we see that there are only four different groups that can be formed of four quantities A,B,C,D by taking three at a time. Thus, the number of combinations of four things taken three at a time is only four.


Note:

It may be observed that the total number of permutations, as given by (1), (2), (3), (4), may be supposed to have been obtained by either :
(1) forming all possible different groups and then rearranging the constituents of each of these groups in different orders in all possible ways; or
(2) filling up three places by means of the four quantities A,B,C,D in all possible ways. If there are m ways of doing a thing and n ways of doing a second thing and p ways of doing a third thing, then the total number of “distinct” ways of doing all these together is \(m \times n \times p.\)

Meaning of factorial :

Factorial of a number (whole number only) is equal to the product of all the natural numbers up to that number. Factorial of n is written as \(\lfloor n\) or \(n!\) and is read as factorial n. Hence :

\(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
\(n! = n (n – 1) (n – 2) \times ...... 3 \times 2 \times 1\)

NOTE : 

(1). \(0! = 1\) (by definition)

(2). \(nP_r = \frac{n! }{ (n – r)!}\) where \(r ≤ n\)

(3). \(C^n_r = \frac{n! }{ [(n – r)! r!]}\) where \(r ≤ n\)



PERMUTATIONS

1. Permutations of n different things taken ‘r’ at a time is denoted by \(P^n_r\) and is given by \(P^n_r\)  \(= \frac{n! }{ (n – r)!}\)

2. The total number of arrangements of n things taken r at a time, in which a particular things always occurs \(= r \times P^{n-1}_{r – 1}\)

3. The total number of permutations of n different things taken r at a time in which a particular thing never occurs \(= P^{n-1}_r\)

4. The total number of permutations of n dissimilar things taken r at a time with repetitions \(= n^r\)

5. 
(a) No. of circular permutations of n things taken all at a time \(= (n – 1)!\)
(b) No. of circular permutations of n different things taken r at a time \(= \frac{P^n_r}{r}\\
\)

7. The number of permutations when things are not all different : If there be n things, p of them of one kind, q of another kind, r of still another kind and so on, then the total number of permutations is given by \(\frac{n! }{ (p! q! r!...)}\)



COMBINATION

1. Number of combinations of n dissimilar things taken ‘r’ at a time is denoted by \(C^n_r\) & is given by \(C^n_r= \frac{n! }{ [ (n – r)! r! ]}\)

2. Number of combinations of n different things taken r at a time in which p particular things will always occur is \(C^{n-p}_{r – p}\\
\)

3. No. of combinations of n dissimilar things taken ‘r’ at a time in which ‘p’ particular things will never occur is \(C^{n-p}_r \)

4. \(C^n_r = C^n_{n – r}\)




The SLOT Method


Another perfectly valid way to complete Permutations and Combinations questions is through what's called the "Slot Method."

In fact, this is much like breaking a quadratic through the guess-and-check method, which is vastly easier than plugging it into the textbook Quadratic Equation (and if you're doing that, please stop).

While the Quadratic Equation is great for situations where the guess-and-check method is prohibitively difficult, the other side of that coin is that it's unnecessarily time-consuming for simple questions.

Going back to P&C: for GRE-level P&C questions, you are unlikely to see anything that really requires pulling out these equations, and they might actually just confuse the issue because they make it difficult to understand what's really going on.

This comes down to asking yourself three Core Questions:
--How Many Spaces?
--How Many Choices?
--Does Order Matter?

Let's take an example: We have seven paintings from which to choose, and three places to hang them on the wall.

Question 1: In how many distinct orders can we hang three paintings?

DO NOT ask whether it's P or C yet; just ask the questions, and that will become apparent:

--How many spaces? 3, naturally.

_ _ _

--How many choices? 7 from which we choose.
>here, just multiply counting down from the 7...

7*6*5

--Does order matter?
>we are looking for DISTINCT ORDERS, so YES -- if YES, it is a Permutation.

That means that we do not need to proceed further. Our answer is 7P3 = 210.

Question 2:
How many distinct groups of three paintings can be hung?

DO NOT ask whether it's P or C yet; just ask the questions, and that will become apparent:

--How many spaces? 3, naturally.

_ _ _

--How many choices? 7 from which we choose.
>here, just multiply counting down from the 7...

7*6*5

--Does order matter?
>we are looking for GROUPS OF THREE, wherein the order of the group doesn't matter, so NO. If NO, it is a Combination.
>At this point, we need to divide by the factorial of the group size (3), because we are treating painting abc=acb=bac=bca=cab=cba, or essentially counting one for every six.
--You can say that dividing by the factorial of group size "randomizes" the numbers within the bag.

7*6*5 / 3! = 7*5 = 35

So our answer is 7C3 = 35.

NOTE on Randomization:

Rather than paintings, let's imagine seven different-colored marbles placed into a box that holds three. Obviously, those would be in some particular order. That means that abc is distinct from acb is distinct from bac and so on. That means each of these is counted as different.

Imagine that we pick up the box and shake it. Now we know which three were chosen, but we have no indication of the order that they currently are in. Dividing by the factorial of group size (3! in this case) is essentially like picking up the box and shaking it.

How this discussion relates to those ever-present equations:

For 7P3, you'll see that n = 7 and r = 3. Therefore, if you plug in 7!/(7-3)! = 7*6*5 you'll find yourself where we were at the beginning, just without the extra work.

For 7C3, you'll see that n = 7 and r = 3. Therefore, if you plug in 7!/(3!(7-3)!) = (7*6*5)/3! = 35. Again, less work.

Some people like the equations and that's totally fine. Pick what works for you. It's good to know multiple ways to do things.



The MISSISSIPI Rule


A useful mnemonic device for the n!/(p!q!r!...) case is called "The Mississippi Rule."

It is as follows: if you ask yourself how many ways you can rearrange the letters in the word "Mississippi," not assuming the letters to make any sense, you would get:

total!/(repeats!)

11!/(4!4!2!) = number of ways to organize the letters in "Mississippi"

In this case, the total number of letters is 11. There are four Is, four Ss, and two Ps.

Not recommended to calculate this fully (the answer is 34650); it is primarily useful as a way to remember the principle.

Counting any group that has repeated terms can be done using the Mississippi Rule, so any time you see something in the form:

AABBB = 5!/(2!3!)

or CCCCDDDD = 8!/(4!4!)

etc. you can apply this calculation.

Thanks to PGTLrowanhand  for the two methods above

source of reference : Difference Between Permutation and Combination by Surbhi S

 

­

Fundamental Principles of Counting

Here we shall discuss two fundamental principles viz. principle of addition and principle of multiplication.
These two principles will enable us to understand Permutations and Combinations. In fact these two principles form the base of Permutations and Combinations.


Fundamental Principle of Multiplication

"If there are two jobs such that one of them can be completed in ‘m’ ways, and another one in ‘n’ ways then the two jobs in succession can be done in ‘m X n’ ways." 

Example :- In her class of 10 girls and 8 boys, the teacher has to select 1 girl AND 1 boy. In how many ways can she make her selection?
Here the teacher has to choose the pair of a girl AND a boy

For selecting a boy she has 8 options/ways AND that for a girl 10 options/ways 
For 1st boy ------- any one of the 10 girls ----------- 10 ways
For 2nd boy ------- any one of the 10 girls ----------- 10 ways
For 3rd boy ------- any one of the 10 girls ----------- 10 ways
-------------
------------
For 8th boy ------- any one of the 10 girls ----------- 10 ways
Total number of ways 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 80 ways    OR   10 X 8 = 80 ways.

Remark :- The above principle can be extended for any finite number of jobs.


Fundamental Principle of Addition

"If there are two jobs such that they can be performed independently in ‘m’ and ‘n’ ways respectively, then either of the two jobs can be performed in (m + n) ways."
 
Example :- In her class of 10 girls and 8 boys, the teacher has to select either a girl OR a boy. In how many ways can she make her selection?
Here the teacher has to choose either a girl OR a boy (Only 1 student)

For selecting a boy she has 8 options/ways OR that for a girl 10 options/ways. The first of these can be performed in 8 ways  and the second in 10 ways. 
Therefore, by fundamental principle of addition either of the two jobs can be performed in (8 + 10) ways. Hence the teacher can make the selection of a student in 18 ways.



Examples 1 :- There are 3 candidates for a classical, 5 for a mathematical, and 4 for a natural science scholarship.

I. In How many ways can these scholarships be awarded?

Clearly classical scholarship can be awarded to anyone of the 3 candidates, similarly mathematical and natural science scholarship can be awarded in 5 and 4 ways respectively. So,
Number of ways of awarding three scholarshipsV= 3 X 5 X 4 = 60 -----------------------[ By Fundamental Principle of Multiplication]

II. In How many ways one of these scholarships be awarded?

Number of ways of awarding one of the three scholarships = 3 + 5 + 4 = 12------------------------[ By Fundamental Principle of Addition]


Example 2 :- A room has 6 doors. In how many ways can a person enter the room through one door and come out through a different
Number of ways coming in the room = 6
Number of ways going out of the room = 5  (He/She cannot go from the same door)

By Fundamental Principle of Multiplication--------> Coming in X Going out = 6 X 5 = 30.


Example 3 :- Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin
i) At any one of the 7 floors
ii) At different floors.

Let the five persons be b,c,d,e,f

I)    b can leave the cabin at any of the seven floors. So he has 7 options
Similarly each of c,d,e,f also has 7 options. Thus the total number of ways in which each of the five persons can leave the cabin at any of the seven floors is 
7 X 7 X 7 X 7 X7 = \(7^5\)

II)    b can leave the cabin in 7 ways. c can leave the cabin in 6 ways, since he can not leave at where b left. In the same way d has 5, e has 4, and f has 3 way.
Hence total number of ways = 7 X 6 X 5 X 4 X 3 = 2520


Example 4 :- In how many ways can 3 prizes be distributed among 4 boys, when
i) No boy gets more than one prize ?
ii) A boy may get any number of prizes ?
iii) No boy gets all the prizes ?

I)    The first prize can be given away to any of the 4 boys, hence there are 4 ways to distribute first prize.  
The second prize can be given away to any of the remaining 3 boys because the boy who got the first prize cannot receive second prize. 
Similarly third prize can be given away to any of the remaining 2 boys
Hence total number of ways are 4 X 3 X 2 = 24
Note :- This is same as Arrangement of 4 boys taken 3 at a time in a way 4P3 = 4!/1! = 4! = 24  (More on this under the Head of Permutations later)

II)    First prize to any one of the 4 boys. Similarly second to any one of the 4 boys, and third as well to any one of the 4 boys = 4 X 4 X 4 = 43 = 64

III)    Since any one of the 4 boys may get all the prizes. So, the number of ways in which a boy gets all the 3 prizes = 4
So the number of ways in which a boy does not get all the prizes = 43 – 4 = 60

Permutations (Arrangement) nPr

Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.

For example, if there are three objects namely x,y, and z, 

then the permutations of these objects, taking two at a time, are xy, yx, yz, zy, xz, zx  = Total 6 Permutations.

NOTE :- It should be noted that in permutations the order of arrangement is taken into account; When the order is changed, a different permutation is obtained.


Now Let’s study some theory

Most of us know the popular formula of calculating number of permutations

It is nPr i.e. Number of all permutations all n distinct things taken r at a time (1 ≤ r ≤ n)  = nPr = \(\frac{n!}{(n-r)!}\)

We will try to know how it is. We will try to prove it by a numerical example.

In how many different ways would you arrange 5 persons on 3 chairs 

Here  n= 5 and r= 3

Arranging 5 persons on 3 chairs is same as filling 3 places when we have 5 different things at our disposal.

The first place can be filled in 5 ways -------------(Remember Fundamental Principal of Multiplication) 

Having filled it, there are 4 things left and anyone of these 4 things can be used to fill second place. So the second place can be filled in 4 ways.

Hence by fundamental principal of multiplication, the first two places can be filled in 5 X 4 ways 

When the first two places are filled, there are 3 things left, so that the third place be filled in 3 ways

So the total number of arrangements will be 5 X 4 X 3.

As per the formula of permutation it will be 5P3 = \(\frac{5!}{(5-3)!}\) = \(\frac{5!}{2!}\) = 5 X 4 X 3

Remark :- Continuing in this manner we can say that number of permutations (or arrangements ) of n things taken all at a time will be n!  
e.g. In how many ways can 6 persons stand in a queue? Here n= r=6 so total number of permutations will be 6! = 720



Example 1 :- It is required to seat 4 Women and 5 Men in a row so that the women occupy the even places. How many such arrangements are possible? 
Total Places = 9
Even Places = 4 i.e. 2nd, 4th, 6th, 8th
4 women can be arranged on 4 even places in 4! Ways.
5 men can be arranged on remaining 5 places in 5!

By Fundamental Principle of Multiplication, Total number of arrangements will be 4! X 5! = 24 X 120 = 2880



Example 2 :- Three men have 4 coats, 5 waist coats, and 6 caps. In how many ways can they wear them?

Number of ways in which 3 Men can wear 4 coats = \(\frac{4!}{1!}\) = 4! = 24
Number of ways in which 3 Men can wear 5 waist coats = \(\frac{5!}{2!}\) = 5 X 4 X 3 = 60
Number of ways in which 3 Men can wear 6 caps = \(\frac{6!}{3!}\) = 6 X 5 X 4 = 120

By Fundamental Principle of Multiplication

Arrangement of 4 Coats AND Arrangement of 5 Waist Coats AND Arrangement of 6 Caps

Total number of ways = 24 X 60 X 120 = 172800

NOTE :- It is unlikely that GRE will ask such an question in the real test, since it involves tedious calculations. However the idea behind putting this here is to show how the Fundamental Principle of Counting works. 


Lets modify the above question

Three men have 4 coats, 5 waist coats, and 6 caps. In how many ways can they wear any one type of them?

Number of ways in which 3 Men can wear 4 coats = \(\frac{4!}{1!}\) = 4! = 24
Number of ways in which 3 Men can wear 5 waist coats = \(\frac{5!}{2!}\) = 5 X 4 X 3 = 60
Number of ways in which 3 Men can wear 6 caps = \(\frac{6!}{3!}\) = 6 X 5 X 4 = 120

By Fundamental Principle of Addition

Arrangement of 4 Coats OR Arrangement of 5 Waist Coats OR Arrangement of 6 Caps

Total number of ways = 24 + 60 + 120 = 204





Example 3 :- How many four digit numbers are there with distinct digits
n = 10 i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9      
and r = 4

Total Number of arrangements = \(\frac{10!}{(10-4)!}\)

But these arrangements also include those numbers which have zero (0) at thousand’s place. 
Such numbers are not four digit numbers and hence need to be excluded.
When 0 is fixed at thousand’s place, we have to arrange remaining 9 digits by taking 3 at a time in a way \(\frac{9!}{(9-3)!}\)
Hence total number of four digit numbers = \(\frac{10!}{(10-4)!}\) – \(\frac{9!}{(9-3)!}\) = 5040 – 504 = 4536




Example 4 :- Find the sum of all the numbers that can be formed with the digits 2, 3, 4, and 5 taken all at a time.

Total Number of numbers = 4! = 24

To find the sum of these 24 numbers, we will find the sum of digits at Unit’s, Ten’s, hundred’s, and thousand’s places  in all these numbers.

Keep 2 at unit’s place. _ _ _ 2  
We can fill the remaining places in 3! i.e. 6 ways. 
So there are 6 numbers in which 2 can occur at unit’s place. 
Similarly we can place all other digits at unit’s places and will get 6 different numbers for each digit.
So, total for the digits in the unit’s place in all the numbers = 2 X 3! + 3 X 3! + 4 X 3! + 5 X 3! 
= (2 + 3 + 4 + 5) 3! = 84-----------Sum of Unit’s Places  (NOTE :- Above is an explanation for this direct formula) 
= (2 + 3 + 4 + 5) 3! X 10 = 840-----------Sum of Ten’s Places
= (2 + 3 + 4 + 5) 3! X 100 = 8400-----------Sum of Hundred’s Places
= (2 + 3 + 4 + 5) 3! X 1000 = 84000-----------Sum of Thousand’s Places
Sum = 93324

By other Way

(2 + 3 + 4 + 5) 3! = 84 -------------> 84 (100 + 101 + 102 + 103) = 93324

Next Case :- What if repetition of digits would have been allowed?

Keeping 2 at unit’s place. _ _ _ 2 we would get 4 X 4 X 4 = 64 numbers.
(2 + 3 + 4 + 5) 64 = 896 -------> 896 (100 + 101 + 102 + 103) = 896 (1+10+100+1000) -------> 896 X 1111 = 995456



Example 5 :- There are 6 periods in each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?

5 periods can be arranged in 6 periods in 6P5 ways. Now one period is left and it can be allotted to any one of the 5 subjects. So number of ways in which remaining one period can be arranged is 5.
Total Number of arrangements = 6P5 X 5 = 3600

Permutations under Certain Conditions

Here we shall see permutations where either repetitions of items are allowed or distinction between some of the items is ignored or a particular item occurs in every arrangement etc.

First Some Theory

1) Permutations of n different objects taken r at a time, when a particular object is to be always included in each arrangement is   r(n-1Pr-1)

Objects = n

Places to be filled = r

Condition  = A Specific Object (lets say x) should always be included in each arrangement.

We will place this x at first place. Now we are left with n-1 objects and r-1 places.

We can arrange the n-1 objects at r-1 places in n-1Pr-1 ways (NOTE Our X was at 1st place in these arrangements)

So each time we push x one place ahead by one position (i.e. 2nd, 3rd,4th,…….rth), we will get n-1Pr-1 new ways of arranging n-1 at r-1 places.

How many places can x go ?   r places right!

So Total Permutations = n-1Pr-1  +  n-1Pr-1   +  n-1Pr-1 …………r times

= r(n-1Pr-1)

2) Permutations of n different objects taken r at a time, when a particular object is never taken in each arrangement is  n-1Pr

Since a particular object is never taken. So, we have to determine the number of ways in which r places can be filled with n-1 distinct objects.



Example 1 :- Make all arrangement of letters of the word PENCIL so that 
i) N is always next to E
ii) N and E are always together.

I)    Let’s keep EN together and consider it one letter. Now we have 5 letters which can be arranged in a row in 5P5 = 5! = 120 ways.

II)    Solve as above. Just keep in mind that now E and N can interchange their places in 2! Ways. So total arrangements = 5!2! = 240 



Example 2 :- A Tea Party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. 4 Persons wish to sit on one particular and 2 on the other side. How many arrangements are possible?

Let the Sides be A and B

Let 4 persons wish to sit at side A. They can sit in 8P4 ways.

Similarly 2 Persons can sit on other side in 8P2

Now we are left with 10 places and 10 persons. So they can be arranged in 10! Ways.

Total Number of Arrangements = 8P4 X 8P2 X 10!

Permutations of Objects not all distinct \(\frac{n!}{p!q!}\)

Many of us may be familiar and adept in solving problems pertaining to this concept

For example if I ask, How many words (with or without meaning) can be formed using all digits of the word INDIA

Almost everybody will say it is \(\frac{5!}{2!}\)  

Why we need to do this division? 

We will first look the underlying Theorem 

Theorem :- The number of mutually distinguishable permutations of n things, taken all at a time, of which p are alike of one kind, q are alike of second such that p + q = n, is  \(\frac{n!}{p!q!}\)

Let’s say n is such that n=p+q where p is set of p alike things and q is set of q alike things

Lets x is the total permutations of n

Let n=5, p=3, q=2 -----------------let’s consider the integers 3,3,3,2,2  where n=33322, p=333, q=22

Now Replace p alike things by p distinct things--------------- Replace integers 3,3,3 by integers 4,5,6  

These replaced integers 4,5,6 could have been permuted themselves in p! or 3! or 6 ways.

Similarly Replace q alike things by q distinct things--------------- Replace integers 2,2 by integers 7,8

These replaced integers 7,8 could have been permuted themselves in q! or 2! or 2 ways.

Now we know x is the total number of permutations of n. 
We also know number of permutations of p is 1 and that of q is also 1. (Number of permutations of 3, 3, 3 = 1  or of 2, 2 = 1)

If all the terms in set P and Q were different the total number of permutations of n would have been 
x X p! X q! = n! --------> x X 3! X 2! = 5! -----------> x X 6 X 2 = 120 ---------x X 12 = 120 

So x = \(\frac{n!}{p!q!}\) ---------- x = \(\frac{120}{12}\) ------> x=10

REMARK :- The total number of permutations of n things, of which p are alike of one kind, q are alike of second kind 
and remaining all are distinct, is  \(\frac{n!}{p!q!}\)


Example 1 :- How many words would you form with the letters of the word MISSISSIPPI  
Total =11, S = 4 times, I = 4 times, P = 2 times
So \(\frac{11!}{4!4!2!}\)  = 34650

Example 2 :- APPLE
Total = 5, P = 2 times, A = 1 time, L = 1 time, E = 1 time
\(\frac{5!}{2!1!1!1!}\) = \(\frac{5!}{2!}\) = 60

Circular Permutations (n-1)!

So far we have discussed linear permutations.  The basic difference between linear permutations and circulations is that every linear arrangement has a beginning and an end, but there is nothing like beginning or end in a circular permutation. Thus, in circular permutations, we consider one object as fixed and the remaining objects are arranged as in case of linear arrangement. 

Thus Total number of circular arrangement of n distinct objects is (n-1)!


Note :- There are certain arrangements in which clockwise and anticlockwise  arrangements are not distinct. e.g. arrangements of beads in a neckless, arrangement of flowers in a garland etc. In such cases number of circular permutations of n distinct objects is  ½ [(n-1)!]


Example :- 
i) Arrange 5 persons around a circular table

(5-1)! = 4! = 24

ii) In how many of these arrangements will two particular persons be next to each other?

Consider two particular persons as one person. We have 4 persons in all. These 4 persons can be seated around a circular table in (4-1)! = 3! Ways. But two Particular persons can be arranged between themselves in 2! Ways
So Total number or arrangements = 3! X 2! = 12
  

Example 2 :- Solve this question in light of above explanation. OA and OE can be found in spoiler.
If 20 persons were invited for a party, in how many ways will two particular persons be seated on either side of the host?
A)    19! X 2
B)    18! X 3!
C)    18! X 2
D)    20!/2!
E)    21!/2!



Example 3 :- In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?

The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3! X 4! = 144

Combinations (Selection) nCr

So far we have discussed Permutations (arrangements) of a certain number or objects by taking some of them or all at a time. 

Most of the times, we are only interested in selection of objects and not their arrangements. In other words, we do not want to specify the ordering of selected objects. 
For example, a company wants to select 3 persons out of 10 applicants, a student wants to choose three books from his library at a time etc.

Suppose there are three objects namely x,y, and z.

We are asked to calculate the permutations (arrangements) of these objects taking 2 at a time 
xy, yx, yz, zy, xz, zx  = Total 6 Permutations.

Now we are asked to calculate the combinations (selections) of these objects taking 2 at a time 
xy, yz, xz  = Total 3 Combinations.

Note the important difference here. In later case we did not differ between xy and yx, as we did in the first case. This is the only difference between Permutations and Combinations.

In Combinations we find different ways of choosing r objects from n given objects 

While in Permutations we find different ways of choosing r objects from n given objects and ways of arranging these r objects.


The formula of permutations(nPr) itself says first selection (nCr) and then arrangement (r!)

 nPr = nCr X r!

\(\frac{n!}{(n-r)!}\) = \(\frac{n!}{r!(n-r)!}\) X r! -----------> \(\frac{n!}{(n-r)!}\) = \(\frac{n!}{(n-r)!}\)

Difference between A Permutation and Combination.

1. In a combination, the ordering of the selected objects is immaterial, whereas in a permutation, the ordering is essential. For example a,b and b,a are same as combinations but different as permutations.

2. Each Combination Correspond to many permutations. For example, the six permutations ABC, ACB, BCA, BAC, CBA, and CAB correspond to the same combination ABC

3. Generally we use the word ‘arrangements’ for permutations and the word ‘selections’ for combinations.



Example 1 :- From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done?
32C4 = \(\frac{32!}{28!4!}\)



Example 2 :- 3 Gentlemen and 3 Ladies are candidates for 2 vacancies. A voter has to vote for 2 candidates. In how many ways can one cast her/his vote.

In all there are 6 candidates and a voter has to vote for any 2 of them. So he can select 2 candidates from 6 candidates in 6C2 ways = \(\frac{6!}{4!2!}\) = \(\frac{(6 X 5)}{2}\) = 5 X 3 = 15



Example 3 :- In how many ways can a cricket eleven be chosen out of a batch of 15 players if

i) There is no restriction on the selection = 15C11

ii) A Particular Player is always chosen = 14C10

iii) A Particular Player is never chosen = 14C11



Example 4 :- Out of 2 Women and 5 Men, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included?

A Committee can be formed in the following ways

1 woman AND 2 men OR 2 women AND 1 man

2C1 X 5C2 + 2C2 X 5C1 = 20 + 5 = 25

Example 5 :- (Pay special attention here, because things to be arranged are identical) In how many ways can 7 plus(+) signs and 5 minus (-) signs be arranged in a row so that no two minus signs are together?

The Plus signs can be arranged in only one way, because all are identical.

|  |+|  |+|  |+|  |+|  |+|  |+|  |+|  |    

There are 8 boxes available for placing 5 minus signs. So 5 boxes can be chosen out of 8 boxes in 8C5 ways.

However since 5 minus signs are identical, they can be placed in 5 chosen boxes in only one way

So Total number of possible arrangements = 1 X 8C5 X 1 = 56
 

Other Important Concepts in Permutations and Combinations

1)\(2^n\) = Selection of any number of things out of n distinct things.

Let there be n distinct things and you have told to choose any number things out of n things

each thing you can either select or deselect. 

So you get two options for dealing with each thing.

There are n things and you have 2 options for each thing.

In all you will have 2 X 2 X 2 X 2 X 2 X……………………n times options  -------By Fundamental Principle of Multiplication 

So there will be \(2^n\) options with you for selection

Example :- Number of ways Shreya can or cannot eat sweets at a party out of 10 distinct sweets available at party = \(2^10\)

Corollary = \(2^n\) – 1 = When zero selections are not allowed. i.e. If Shreya has asked to select at least one sweet.

Note :- This is especially important in SET concept.  

For example let the Set A = (1,2,3) 

Question :- How many subsets does the Set A have.

Solution :- We know a subset can be formed by taking any number of elements (or even zero) from its superset.  
\(2^n\) stands for selection of any number of things out of given things. 

So the number of subsets of Set A will be \(2^3\) = 8 (Here n=3 because Set A have 3 distinct elements)
(0) (1) (2) (3) (1,2) (2,3) (1,3) (1,2,3)


2) n+1 = Selection of any number of things out of n identical things

Selection of any number of balls from 3 red balls.

Zero ball selected – 1 way

One ball selected – 1 way

Two balls selected – 1 way

Three balls selected – 1 way

Total (3 +1 ) = 4 ways.


3) \(\frac{n(n-3)}{2}\) = Number of diagonals of n sided polygon


4) nC2 - rC2 + 1 =  Number of straight lines formed by n points of which r are collinear


5) nC3 - rC3 = Number of triangles formed by n points of which r are collinear


6) With m parallel lines intersected by n parallel lines \(\frac{mn(m-1)(n-1)}{4}\) parallelograms can be formed. 
Note :- This formula is derived from mC2 X nC2


7) Sum of the Numbers formed by using n digits taking all at a time is (n-1)! (1111…..n times) (Sum of Digits)


8) If n is even, then nCr will be maximum for r = \(\frac{n}{2}\)

Special Thanks to

1) Prof. Dr. R. D. Sharma

B.Sc. (Hons) (Gold Medalist), M.Sc. (Gold Medalist), Ph.D.  
Author of CBSE Math Books

2) Mr. Rupesh Mishra (IIT, Mumbai) - My P&C Guru

Regards,­

­
Probability of an event occurring \(= \frac{Number of favourable outcomes }{Number of all possible outcomes}\)

Note :

(a). If an event E is sure to occur, we say that the probability of the event E is equal to 1 and we write \(P (E) = 1\).

(b). If an event E is sure not to occur, we say that the probability of the event E is equal to 0 and we write \(P (E) = 0\). Therefore for any event E, \(0 ≤ P (E) ≤ 1\)


Mathematical definition of probability :

(A) If the outcome of an operation can occur in n equally like ways, and if m of these ways are favorable to an event E, the probability of E, denoted by \(P (E)\) is given by \(P (E) = \frac{m}{n}\)

(B) As \(0 ≤ m ≤ n\), therefore for any event E, we have \(0 ≤ P (E) ≤ 1\)

(C) The probability of E not occuring, denoted by P(not E), is given by \(P(not E)\) or \( P(E') = 1 – P(E)\)

(D) Odds in favour \(=\frac{ No. of favourable cases }{ No. of unfavorable cases}\)

(E) Odds against \(= \frac{No. of unfavourable cases }{ No. of favorable cases}\)



Mutually Exclusive Events :

Two events are mutually exclusive if one happens, the other can’t happen & vice versa. In other words, the events have no common outcomes. For example

(a). In rolling a die

E :– The event that the no. is odd
F :– The event that the no. is even
G :– The event that the no. is a multiple of three

(b). In drawing a card from a deck of 52 cards

E :– The event that it is a spade
F :– The event that it is a club
G :– The event that it is a king

In the above 2 cases events E & F are mutually exclusive but the events E & G are not mutually exclusive or disjoint since they may have common outcomes.


Addition law of Probability :

If E & F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by : P(E or F) = P(E) + P(F)

If the events are not mutually exclusive, then P(E or F) = P(E) + P(F) – P(E & F together).

Note : 

Compare this with n(A ∪ B) = n(A) + n(B) – n(A ∩ B) of set theory. Similarly P (neither E nor F) = 1 – P(E or F).


Independent Events:

Two events are independent if the happening of one has no effect on the happening of the other.

Example :

1.On rolling a die & tossing a coin together
E :– The event that no. 6 turns up.
F :– The event that head turns up.

2.In shooting a target
E :– Event that the first trial is missed.
F :– Event that the second trial is missed.

In both these cases events E & F are independent. BUT 3. In drawing a card from a well-shuffled pack

E :– Event that first card is drawn
F :– Event that second card is drawn without replacing the first
G :– Event that second card is drawn after replacing the first

In this case E & F are not Independent but E & G are independent.


Multiplication Law of Probability:

If the events E & F are independent then P(E & F) = P (E) x P (F) & P (not E & F) = 1 – P (E & F together)

­

How To Solve: Probability Problems involving Coin Toss

Attached pdf of this Article as SPOILER at the top! Happy learning!

Hi All,

I have recently uploaded a video on YouTube to discuss Probability Problems involving Coin Toss in Detail:




Following is covered in the video

¤ What is Probability of an Event ?
¤ Probability: Tossing 1 Fair Coin
¤ Probability: Tossing 2 Fair Coins
¤ Probability: Tossing 3 Fair Coins

Theory

What is Probability of an Event?

• Probability of an Event is the Likelihood of occurrence of that event.

• Probability that an Event, E, will occur is denoted by P(E)

P(E) = "No. of successful Outcomes" /"Total number of Outcomes"


Probability: Tossing 1 Fair Coin

Fair coin is a coin which has equal probability of getting a Head or a Tail.

When we toss one coin or when we toss a coin one time then 

• Total Number of Outcomes = 2 ( We can get a Head(H) or a Tail(T) )

• Outcomes are { H , T }

• Probability of Getting a Head, P(H) = \(\frac{1}{2}\) (As there are two outcomes and only one out of those results in a head)

• Probability of Getting a Tail, P(T) = \(\frac{1}{2}\) = (As there are two outcomes and only one out of those results in a Tail)


Probability: Tossing 2 Fair Coins

When we toss two coins or when we toss a coin two times then 

• Total Number of Outcomes = \(2^2\) = 4 ( In each of the toss we can get a Head(H) or a Tail(T) => 2*2 = 4 )

• Outcomes are { HH, HT, TH, TT }

• Probability of Getting 0 Head, P(0H) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 0 Head or 2 Tails)

• Probability of Getting 1 Head, P(1H) = \(\frac{2}{4}\) = \(\frac{1}{2}\) (As there are two outcomes out of 4 where we get 1 Head i.e. HT, TH)

• Probability of Getting 2 Head, P(2H) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 2 Head i.e. HH)

• Probability of Getting 0 Head, P(0T) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 0 Tail or 2 Heads)

• Probability of Getting 1 Head, P(1T) = \(\frac{2}{4}\) = \(\frac{1}{2}\) (As there are two outcomes out of 4 where we get 1 Tail i.e. HT, TH)

• Probability of Getting 2 Head, P(2T) = \(\frac{1}{4}\) (As there is ONLY one outcome out of 4 where we get 2 Tails i.e. TT)

• P(0H) + P(1H) + P(2H) = \(\frac{1}{4}\) + \(\frac{2}{4}\) + \(\frac{1}{4}\) = 1

• Probability of at least 1 Head = P(1H) + P(2H) = \(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)

• Probability of at least 1 Head = 1 - P(0H) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

• Probability of at least 1 Tail = 1 - P(0T) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)



Probability: Tossing 3 Fair Coins

When we toss three coins or when we toss a coin three times then 

• Total Number of Outcomes = \(2^3\) = 8 ( In each of the toss we can get a Head(H) or a Tail(T) => 2*2*2 = 8 )

• Outcomes are { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

• Probability of Getting 0 Head, P(0H) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 0 Head or 3 Tails)

• Probability of Getting 1 Head, P(1H) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 1 Head i.e. HTT, THT, TTH)
We can also find this by finding the position of that one Head out of three slots in 3C1 = \(\frac{3!}{1!*2!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 2 Head, P(2H) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 2 Heads i.e. HHT, THH, HTH)
We can also find this by finding the position of those 2 Heads or 1 Tail out of three slots in 3C2 = \(\frac{3!}{2!*1!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 3 Head, P(3H) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 3 Head i.e. HHH)

• Probability of Getting 0 Tail, P(0T) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 0 Tail or 3 Heads)

• Probability of Getting 1 Tail, P(1T) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 1 Tail i.e. THH, HTH, HHT)
We can also find this by finding the position of that one Tail out of three slots in 3C1 = \(\frac{3!}{1!*2!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 2 Tail, P(2T) = \(\frac{3}{8}\) (As there are three outcomes out of 8 where we get 2 Tails i.e. TTH, THT, HTT)
We can also find this by finding the position of those 2 Tails or 1 Head out of three slots in 3C2 = \(\frac{3!}{2!*1!}\) = 3 ways and divide it by the total number of outcomes which is 8

• Probability of Getting 3 Tail, P(3T) = \(\frac{1}{8}\) (As there is ONLY one outcome out of 8 where we get 3 Tail i.e. TTT)

• P(0H) + P(1H) + P(2H) + P(3H) = \(\frac{1}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{1}{8}\) = 1

• Probability of at least 1 Head = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

• Probability of at least 2 Head = P(2H) + P(3H) = \(\frac{3}{8}\) + \(\frac{1}{8}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

• Probability of at least 1 Tail = 1 - P(0T) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

• Probability of at least 2 Tail = P(2T) + P(3T) = \(\frac{3}{8}\) + \(\frac{1}{8}\) = \(\frac{4}{8}\) = \(\frac{1}{2}\)­

­

How To Solve: Probability Problems involving Rolling a Dice

Attached pdf of this Article as SPOILER at the top! Happy learning!

Hi All,

I have recently uploaded a video on YouTube to discuss Probability Problems involving Rolling a Dice in Detail:

[youtube]https://www.youtube.com/watch?v=5PVZYQXet18[/youtube]


Following is covered in the video

¤ What is Probability of an Event ?
¤ Probability: Rolling 1 Fair Dice
¤ Probability: Rolling 2 Fair Dice
¤ Example Problems on Rolling 2 Fair Dice

Theory

What is Probability of an Event?

• Probability of an Event is the Likelihood of occurrence of that event.

• Probability that an Event, E, will occur is denoted by P(E)

P(E) = "No. of successful Outcomes" / "Total number of Outcomes"


Probability: Rolling 1 Fair Dice

Fair dice is a dice which has equal probability of getting any of the six numbers.

When we roll one dice or when we roll a dice one time then 

• Total Number of Outcomes = 6 ( We can get any number from 1 to 6 )

• Outcomes are { 1, 2, 3, 4, 5, 6 }

• Probability of Getting a one, P(1) = \(\frac{1}{6}\) (As there are is ONLY one way out of 6 in which we can get a 1)

• Similarly, P(2) = P(3) = P(4) = P(5) = P(6) = \(\frac{1}{6}\)

• Probability of Getting a number greater than 3, P(>3) = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As there are 3 numbers greater than out of 6 i.e. 4, 5 and 6)

• P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = Total Probability = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = 1

• Probability of Getting a number greater than 3, P(>3) = P(4) + P(5) + P(6) = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

• Probability of Getting an Odd number, P(Odd) = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As there are 3 odd numbers out of the 6 numbers i.e. 1, 3 and 5)

• Probability of Getting an Even number, P(Even) = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As there are 3 even numbers out of the 6 numbers i.e. 2, 4 and 6)

• Probability of Getting a Prime number, P(Prime) = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As there are 3 prime numbers out of the 6 numbers i.e. 2, 3 and 5)


Probability: Rolling 2 Fair Dice

When we roll two dice or when we roll a dice two times then 

• Total Number of Outcomes = \(6^2\) = 36 ( As we can get any number from 1 to 6 in each of the two rolls => 6 * 6 = 36)

• Outcomes are

(1,1) ,(1,2) ,(1,3), (1,4), (1,5), (1,6)
(2,1) ,(2,2) ,(2,3), (2,4), (2,5), (2,6)
(3,1) ,(3,2) ,(3,3), (3,4), (3,5), (3,6)
(4,1) ,(4,2) ,(4,3), (4,4), (4,5), (4,6)
(5,1) ,(5,2) ,(5,3), (5,4), (5,5), (5,6)
(6,1) ,(6,2) ,(6,3), (6,4), (6,5), (6,6)

In each of the two places _ _ any number from 1 to 6 can come.


Example Problems on Rolling 2 Fair Dice

Q1. Find the probability that both the outcomes are Odd, P(Both outcomes are Odd)?

Solution: We have two places _ _ , in each of the two places any of the three odd numbers (1, 3, 5) can come.
=> We have 3 * 3 = 9 ways possible
=> P(Both outcomes are Odd) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

Q2. Find the probability that both the outcomes are Even, P(Both outcomes are Even)?

Solution: We have two places _ _ , in each of the two places any of the three even numbers (2, 4, 6) can come.
=> We have 3 * 3 = 9 ways possible
=> P(Both outcomes are Even) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

Q3. Find the probability that both the outcomes are Prime, P(Both outcomes are Prime)?

Solution: We have two places _ _ , in each of the two places any of the three Prime numbers (2, 3, 5) can come.
=> We have 3 * 3 = 9 ways possible
=> P(Both outcomes are Prime) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

Q4. Find the probability that sum of the two outcomes is Odd, P(Sum of the two outcomes in Odd)?

Solution: Given that it is a fair dice so there is an equal probability of getting the sum as odd or even
=> P(Sum of the two outcomes in Odd) = \(\frac{1}{2}\)

Let's solve using one more method
We have two places _ _  and for the sum to be odd, one of them has to be odd and other has to be even
Odd Even -> Number of cases = 3 (for first odd) * 3 (for second even) = 9
Even Odd -> Number of cases = 3 (for first even) * 3 (for second odd) = 9
Total cases = 9 + 9 = 18
=> P(Sum of the two outcomes in Odd) = \(\frac{18}{36}\) = \(\frac{1}{2}\)

Q5. Find the probability that sum of the two outcomes is Even, P(Sum of the two outcomes in Even)?

Solution: Given that it is a fair dice so there is an equal probability of getting the sum as odd or even
=> P(Sum of the two outcomes in Even) = \(\frac{1}{2}\)

Let's solve using one more method
We have two places _ _  and for the sum to be odd, either both of them should be odd or both should be even
Odd Odd -> Number of cases = 3 (for first odd) * 3 (for second odd) = 9
Even Even -> Number of cases = 3 (for first even) * 3 (for second even) = 9
Total cases = 9 + 9 = 18
=> P(Sum of the two outcomes in Even) = \(\frac{18}{36}\) = \(\frac{1}{2}\)

Q6. Find the probability that sum of the two outcomes is a prime number, P(Sum of the two outcomes is Prime)?

Solution: We know that between 2 (1+1) and 12 (6+6) we have 2, 3, 5, 7, 11 as the prime numbers.

Let's list down the cases in which sum of the two outcomes will be a Prime number.
(1,1), (1,2), (1,4), (1,6)
(2,1), (2,3), (2,5)
(3,2), (3,4)
(4,1), (4,3)
(5,2), (5,6)
(6,1), (6,5)

=> Total cases = 15
=> P(Sum of the two outcomes is Prime) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

Q7. Find the probability that sum of the two outcomes is a multiple of 3, P(Sum of the two outcomes is a multiple of 3)?

Solution: Let's list down the cases in which sum of the two outcomes will be a multiple of 3
(1,2), (1,5)
(2,1), (2,4)
(3,3), (3,6)
(4,2), (4,5)
(5,1), (5,4)
(6,3), (6,6)

=> Total cases = 12
=> P(Sum of the two outcomes is a multiple of 3 = \(\frac{12}{36}\) = \(\frac{1}{3}\)

Q8. Find the probability that sum of the two outcomes is a multiple of 5, P(Sum of the two outcomes is a multiple of 5)?

Solution: Let's list down the cases in which sum of the two outcomes will be a multiple of 5
(1,4)
(2,3)
(3,2)
(4,1), (4,6)
(5,5)
(6,4)

=> Total cases = 7
=> P(Sum of the two outcomes is a multiple of 5) = \(\frac{7}{36}\)

Q9. Find the probability that product of the two outcomes is a prime number, P(Product of the two outcomes is Prime)?

Solution: Let's list down the cases in which product of the two outcomes will be a Prime number.
(1,2), (1,3), (1,5)
(2,1)
(3,1)
(5,1)

=> Total cases = 6
=> P(Product of the two outcomes is Prime) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Q10. Find the probability that product of the two outcomes is a multiple of 3, P(Product of the two outcomes is a multiple of 3)?

Solution: Let's list down the cases in which product of the two outcomes will be a multiple of 3
(1,3), (1,6)
(2,3), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,3), (4,6)
(5,3), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

=> Total cases = 20
=> P(Product of the two outcomes is a multiple of 3 = \(\frac{20}{36}\) = \(\frac{5}{9}\)

Q11. Find the probability that product of the two outcomes is a multiple of 5, P(Product of the two outcomes is a multiple of 5)?

Solution: Let's list down the cases in which product of the two outcomes will be a multiple of 5
(1,5)
(2,5)
(3,5)
(4,5)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,5)

=> Total cases = 11
=> P(Product of the two outcomes is a multiple of 5) = \(\frac{11}{36}\)


Hope it helps!
Good Luck!­

­Break Up the Prompt: General principles 

• Use variable names that represent the question's quantities. Suppose a machine performs a certain task you can jot down using the letter \(m =\) machine or \(t=\) task. At the end, when you set up a suitable equation, you immediately will be able to identify what the \(m\) or \(t\) stands for. If you use variables, such as x is too vague.
  
• Break down the problem into as many pieces as possible. Construct your equation one step at a time. Not try to write the entire situation in one shot, creating a supergiant equation. You will get lost on the way process, probably than not.
  
• When you have already set up the equation/s be sure to understand your final goal. Not only you must solve the equation per see but also you must solve it, understanding what the problem asks or means. Be sure to know EXACTLY what the question wants you to find out.
  
• if you have established the relationship between the two unknowns, label the smallest quantity of the two with the letter you pick.
  
• If the translation is a bit complex, paraphrase the sentences and personalize its words in a way you clearly understand what you are looking for.
  
• Organize step by step your translation of words into math. DO NOT mess up. Keep them clean with an understandable flowing. Do not rush: one of the students' most significant mistakes is running, pressured by the time. You have time. If you evaluate well up-front a problem, then you waste LESS time. 
  
  
• I do not know where or how to start
  
  A passive student or someone who does not attack the question proactively waits for something happens to solve the question. On the other hand, those who aggressively attacks the question try always to find
  
   - a possible relationship.
   - try something, and if this does not lead anywhere, immediately switch in another direction without wasting time.
   - try multiple things simultaneously, seeing what best suits the question to solve in that precise moment. Sometimes, in solving a question, the students still process the previous question that they are unsure about. Focus 100% on the task you do have in front of you. An active thinker identifies unknowns and creates variables to represent them.
  
  
• I don’t know what they want me to do.
  
  If you wait until you know everything to start the price to solve a question, I.E. you wait like so many people do to be perfect, you will never begin. Begin small and figure out what you need for a piece by piece. 
  
  Look at this problem and how we set up and solve the problem step by step.
  
  In 10 years, Bill will be twice as old as Jeffery. Today, Bill is 4 times as old as Jeffery. How old is Bill?
  
  Solution
  
  Today, Bill is 4 times as old as Jeffery
  Let J = Jeffery's PRESENT age
  So, 4J = Bill's PRESENT age
  
  In 10 years, Bill will be twice as old as Jeffery.
  Since we already know each person's PRESENT age, we know the following...
  J + 10 = Jeffery's age IN 10 YEARS
  4J +10 = Bill's age IN 10 YEARS 
  
  If Bill is twice as old as Jeffrey in 10 years we can write: 4J + 10 = 2(J + 10)
  Expand to get: 4J + 10 = 2J + 20
  Solve to get: J = 5
  
  So, Jeffery's PRESENT age is 5
  Since 4J = Bill's PRESENT age, we know that Bill's PRESENT age = 4(5) = 20

Translating Rightly 

Word Translation
Words	Math
A is half the size of B	\(A = \frac{1}{2} B\) NOT \(B =\frac{1}{2}A\)
A is 5 less than B	\(A=B-5\) NOT \(B=5-B\)
A is less than B	\(A < B\) NOT \(A > B\)
Carcass bought twice as many cars as bikes	\(C=2B\) NOT \(2C=B\)
P is X percent of Q	\(P=\frac{X}{100} Q \) NOT \(P=X\)%\(Q\)
Pay $30 per pair of boots, then $28 per additional pair	\(n=#\) the number of pair of boots purchased \(+\)  \(T =\) the total amount paid for $ \(T=$30 + $28 \times (n-1)\)

Most common word problem translations 






More Examples 


1. Find two consecutive odd numbers the difference of whose squares is 296.

Sol. Let the numbers be 2X + 1 and 2X + 3 Then (2X + 3)² – (2X + 1)² = 296 ⇒ X = 36 
Hence 2X+ 1 = 2 x 36 + 1 = 73 and 2X + 3 = 75 ⇒ The required numbers are 73 and 75. [Verification. (75)² – (73)²= 5625 – 5329 = 296]


2. A is 29 years older than B, B is 3 years older than C and D is 2 years younger than C. Two years hence A’s age will be twice the united ages of B, C and D. Find their present ages.

Sol. Let D’s age be = X years Then C’s age = (X + 2) years, B’s age = (X + 5) years and A’s age = (X + 34) years. Two yrs hence A’s, B’s, C’s and D’s ages will be X + 36, X + 7, X + 4 and X + 2 years respectively. ⇒ 2 (X + 2 + X + 4 + X + 7) = X + 36 ⇒ X = 2 ⇒ A’s age = 36 yrs; B’s age = 7 yrs, C’s age = 4 yrs; D’s age = 2 yrs.


3. A number consists of three consecutive digits, that in the unit’s place being the greatest of them three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digit. Find the number.

Sol. Let the hundred’s digit be X. Then the ten’s digit = X + 1 and the unit’s digit = X + 2 ⇒ The number = 100 x X + 10(X + 1) + X + 2 = 111X + 12. The number formed by reversing the digits = 100(X + 2) + 10(X + 1) + X = 111X + 210 ⇒ 111X + 210 – 111X – 12 = 22 (X + 2 + X + 1 + X) ⇒ X = 2. Hence the required number = 234.


4. The crew of a boat can row at the rate of 5 miles an hour in still water. If to row 24 miles, they take 4 times as long as to row the same distance down the river, find the speed at which the river flows.

Sol. Let X miles per hour be the speed of the river. Hence, on equating the times, we get: \(\frac{24}{(5 – X)} = 4 \times \frac{24}{(X + 5)} ⇒ X = 3\) Thus, the river flows at the rate of 3 miles an hour.


5. The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length decreases by 7 m and breadth increases by 5 m. Find the dimensions of the rectangle.

Sol. Let the length of the rectangle be X m and breadth of the rectangle = Y m. Area = XY sq. m.

I Case : Length = (X + 7) m and breadth = (Y – 3) m ⇒ Area = (X + 7) (Y – 3) sq. m. ⇒ (X + 7) (Y – 3) = XY or – 3X + 7Y – 21 = 0 .... 

II Case : Length = (X – 7) m and breadth = (Y + 5) m ⇒ (X – 7) (Y + 5) = XY or 5X – 7Y – 35 = 0 .... (2) ⇒ Y = 15 and X = 28 Hence length = 28 m and breadth = 15 m Answer.

Note : Assume L = 28 and B = 15 as an option and try checking the conditions given in the problem. You will see that working backward is exceptionally fast in such cases.


6. The ratio of incomes of two persons is 9 : 7, and the ratio of their expenditures is 4 : 3. If they save Rs. 200 per month, find their monthly incomes.

Sol. Let the monthly income of the first person be Rs 9X and the monthly income of the second person be Rs 7X. Let the expenditure of the first person be 4Y and the expenditure of the second person be 3Y. ⇒ Saving of the first person = Rs (9X – 4Y) and solving of second person = Rs (7X – 3Y). Using the given informations, we have : 9X – 4Y = 200 .... (1) and 7X – 3Y = 200 .... X = 200. Hence, the monthly income of first-person = Rs. 9 x 200 = Rs. 1800 and the monthly income of second person = Rs. 7 x 200 = Rs. 1400


7. Find two consecutive even numbers such that 1/6th of the greater exceeds 1/10th of the smaller by 29.

Sol. Let the numbers be 2X and 2X + 2 Then (2X + 2)/6 – 2X/10 = 29. ⇒ X = 21 Hence 2X = 430 and 2X + 2 = 432. ⇒ The required numbers are 430 and 432. [Verification. \(\frac{432}{6} – \frac{430}{10} = 72 – 43 = 29\)]


8. A number consists of two digits whose sum is 12. The ten’s digit is three times the unit’s digit. What is the number?

Sol. Let the unit’s digit be X, Then the ten’s digit is 12 – X. ⇒ 3X = 12 – X ⇒ X = 3 Hence the number is 93. [Verification. 9 = 3 x 3; and 9 = 3 = 12]


9. A train traveled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the journey would have occupied 4 hours less; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance.

Sol. Let us suppose that X miles per hour is the speed of the train and Y hours is the time taken for the journey. ⇒ Distance travelled = XY = (X + 6) (Y – 4) = (X – 6) (Y + 6) This gives two simultaneous equations. Solving, we get : X = 30, Y = 24 ⇒ Distance = XY = 720 miles


10. A sum of money was divided equally among a certain number of persons; had there been six more, each would have received a rupee less, and had there been four fewer, each would have received a rupee more than he did; find the sum of money and the number of men.

Sol. Let X be the number of persons and Rs Y be the share of each. Then by the conditions of the problem, we have (X + 6) (Y – 1) = XY .......(1) (X – 4) (Y + 1) = XY .......(2). Thus the number of person is X = 24 and the share of each is Y = Rs 5. The sum of money = 5 x Rs 24 = Rs 120.

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1. Speed: The rate at which anything covers a particular distance is called its speed ⇒ Speed = Distance Travelled / Time Taken 

Generally, speed is expressed in the following units: miles/hr, km/hr, m/sec, m/min, etc.

1. A policeman goes after a thief who is 176 m ahead of him. When and where will the policeman catch the thief when they run at the rates of 11440 and 10560 meters per hour respectively?

Sol. Time to \(\frac{catch }{ overtake} = \frac{lead distance }{ difference of speeds} = \frac{176 }{ (11440 – 10560) }= \frac{176 }{ 880} = \frac{1}{5} hours = 12\) minutes ⇒ The time required to overtake the thief = 12 min.

(b) The distance from the starting point \(= \frac{11440 \times 12}{60}\) kms = 2288 meters


2. If A goes from X to Y at U km/hr and comes back from Y to X at V km/hr, then Average speed during the whole journey \(= \frac{2UV }{ (U + V)}\) km/hr.

2. If I walk at the rate of 4 kms an hour, I reach my destination 30 min too late; If I walk at the rate of 5 kms an hour I reach 30 minutes too soon. How far is my destination ?

Sol. Let time taken be T hrs for the distance to be covered at the normal speed (neither fast nor slow). Then we have 4 (T + 0.5) = 5 (T – 0.5) {Note : 0.5 here is 30 min} ⇒ T = 4.5 hours ⇒ Distance \(= 4 (T + 0.5) = 4 \times 5 = 20\) km.


3. If a man changes his speed in the ratio m : n then the ratio of times taken becomes n : m.

3. A man rows 18 kms down a river in 4 hours with the stream and returns in 12 hours; find his speed and also the velocity of the stream.

Sol. Speed with the stream \(= \frac{18}{4} = 4.5\) kms an hour. ⇒ Speed against the stream \(= \frac{18}{12} = 1.5\) kms an hour. ⇒ Speed of the stream \(= \frac{1}{2} (4.5 – 1.5) = 1.5\) kms an hour and his speed = 4.5 – 1.5 = 3 kms an hour


4. If three men cover the same distance with speeds in the ratio a : b : c, the times taken by these three will be resp. in the ratio \(\frac{1}{a} : \frac{1}{b} : \frac{1}{c}\).

4. A, B and C can walk at the rates of 3, 4, 5 kms an hour. They start from X at 1, 2, 3 o’clock respectively; when B catches up with A, B sends him back with a message to C; when will C get the message ?

Sol. In one hour A covers 3 kms Now (B – A) = 1 kmph hence B catches up A after 3 hours i.e. at 5 o’ clock. Now upto 5 o’ clock A has covered 12 kms and C has covered 10 kms. Hence Distance between A & C = 2 kms and their relative speed (3 + 5) = 8 kmph. To cover 2 kms at 8 kmph, Time \(=\frac{ 2}{8}\) hours = 15 min. Hence C gets the message at 15 minutes past 5 o’clock.


Relative Speed:

When two objects travel in the same direction, relative speed = difference of speeds

When two objects travel in opposite directions, relative speed = sum of speeds


5. When two objects travel in the same direction, relative speed = difference of speeds time to catch / overtake = lead distance / difference of speeds

5. A student walks to school at the rate of 2.5 kms an hour and reaches 6 minutes too late. Next day he increases his speed by 2 kms an hour and then reaches there 10 minutes too soon. Find the distance of the school from his home.

Sol. Let t be the usual time We have \(⇒ 2.5 \times (t + \frac{1}{10}) = 4.5 (t – \frac{1}{6})\), or \(t = \frac{1}{2}\) hours. Hence distance \(= 2.5 (\frac{1}{2} + \frac{1}{10}) = 2.5 \times \frac{6 }{10} = 1.5\) km.


6. When two objects travel in the Opposite directions, relative speed = sum of speeds time to meet = lead distance / sum of speeds.

6. A man can row in still water a distance of 4 kms in 20 minutes and 4 kms with the current in 16 min. How long will it take him to row the same distance against the current ?

Sol. \(X = \frac{4}{(\frac{20}{60})} = 12\), \(X + Y = 4 \times \frac{60 }{16} = 15\) or \(Y = 3\). or Time \(= \frac{4 }{ (X – Y)} = 4 \times \frac{60 }{9} = \frac{80}{3}\) minutes


7. If the speed of a boat (or man) in still water be \(X \frac{km}{hr}\), and the speed of the stream (or current) be \(Y \frac{km}{hr}\), then

(a) Speed of boat with the stream (or Downstream or D/S) \(= \frac{(X + Y) km}{hr}\)

(b) Speed of boat against the stream (or upstream or U/S) \(= \frac{(X – Y) km}{hr}\)

We have \(X = \frac{[(X + Y) + (X – Y)] }{ 2}\) and \(Y = \frac{[(X + Y) – (X – Y)] }{ 2} ⇒\) Boat’s speed in still water \(= \frac{[Speed downstream + Speed upstream] }{ 2}\) Speed of current \(= \frac{[Speed downstream – Speed upstream] }{ 2}\)

­1. If a man can do a piece of work in N days (or hours or any other unit of time), then the work done by him in one day will be \(\frac{1}{N}\) of the total work.

1. If A and B together finish a piece of work in 10 days & B alone can finish it in 20 days. In how many days can A alone finish the work ?

Sol. Let X and Y be the number of days required by A and B respectively. ⇒ By the standard formula, \(\frac{XY}{(X + Y)} = 10\) & \(Y = 20 ⇒ \frac{X \times 20 }{(X + 20)} = 10\) or X = 20 days 


2. If A is twice as good a workman as B, then A will take half the time B takes to finish a piece of work.

2. Four men working together all day, can finish a piece of work in 11 days; but two of them having other engagements can work only one half–time and quarter time respectively. How long will it take them to complete the work ?

Sol. Each man will take 11 x 4 = 44 days to complete the work. If one man works half day/day he will take 44 x 2 = 88 days to finish the work. Similarly, a man working quarter \(\frac{day}{day}\) will take 44 x 4 = 176 days to finish the work. When these work together they will require \(\frac{1 }{\left[ \begin{array}{cc|r} (\frac{1}{44}) + (\frac{1}{44}) + (\frac{1}{88}) + (\frac{1}{176}) \end{array} \right] }= 16\) days.


3. If A and B can do a piece of work in X & Y days respectively while working alone, they will together take \(\frac{XY }{ (X + Y)}\) days to complete it.

3. 20 men can complete a piece of work in 10 days, but after every 4 days 5 men are called off, in what time will the work be finished ?

Total work = 20 x 10 = 200 monday
1. First 4 days' work = 20 x 4 = 80 md
2. Next 4 days' work = 15 x 4 = 60 md
3. Next 4 days' work = 10 x 4 = 40 md
4. Next 4 days' work = 5 x 4 = 20

\(Σ = 200\) md

Hence, days reqd = 4 + 4 + 4 + 4 = 16



4. If A, B, C can do a piece of work in X, Y, Z days respectively while working alone, they will together take \(\frac{XYZ }{ [XY + YZ + ZX]}\) days to finish it.

4. A vessel can be filled by one pipe A in 10 minutes, by a second pipe B in 15 minutes. It can be emptied by a waste pipe C in 9 minutes. In what time will the vessel be filled if all the three were turned on at once ?

Sol. We will follow exactly the same method as in time & work. The part of a vessel filled in 1 minute when all three are on \(= \frac{1}{10} + \frac{1}{15} – \frac{1}{9} = \frac{1}{18} ⇒\) Total vessel will be filled in 18 minutes


5. If an inlet pipe can fill a cistern in X hours, the part filled in 1 hour = \(\frac{1}{X}\)

5. Three pipes A, B and C can fill a cistern in 15, 20 and 30 min resp. They were all turned on at the same time. After 5 minutes the first two pipes were turned off. In what time will the cistern be filled ?

Sol. A, B and C can fill \((\frac{1}{15} + \frac{1}{20} + \frac{1}{30})\) or \(\frac{3}{20}\) of the cistern in 1 minute ⇒ A, B and C filled \((\frac{3}{20} x 5)\) or \(\frac{3}{4}\) of the cistern in 5 min. Now A and B are turned off \(⇒ (1 – \frac{3}{4})\) or \(\frac{1}{4}\) of the cistern wil be filled by C ⇒ C will fill \(\frac{1}{4}\) of the cistern in \((30 \times \frac{1}{4})\) or 7.5 minutes ⇒ The cistern will be filled in 7.5 + 5 or 12.5 min


6. If an inlet pipe can fill a tank in X hours and an outlet pipe empties the full tank in Y hours, then the net part filled in 1 hour when both the pipes are opened \(= (\frac{1}{X}) – (\frac{1}{Y}) ⇒\) In 1 hour, the part filled (or emptied) \(= \frac{1}{X }– \frac{1}{Y} ⇒ \) Time required to fill or empty the tank \(= \frac{XY }{ (X \sim Y)}\) hours. \(X \sim Y\) indicates \([X – Y]\) or \([Y – X]\), whichever is positive).

6. A cistern can be filled by two taps A and B in 12 minutes and 14 minutes respectively and can be emptied by a third in 8 minutes. If all the taps are turned on at the same moment, what part of the cistern will remain unfilled at the end of 7 minutes ?

Sol. We have \((\frac{7}{12}) + (\frac{7}{14}) – \frac{7}{8} = \frac{5}{24}\) part filled in 7 minutes. Hence \(1 – \frac{5}{24} = \frac{19}{24}\) th of the tank is unfilled.