elinka wrote:
sorry, still dont get it .. whats 4C2 .. 5C3 .. how do u calculate that?
To deal with such question you should be aware of the combinatorics or fundamental principles of counting. So below is the short review:
FUNDAMENTAL PRINCIPLE OF COUNTINGPrinciple of MultiplicationIf an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘mxn’ ways.
Principle of AdditionIf an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then either of these two operations can be performed in ‘m+n’ ways.(provided only one has to be done) FACTORIAL ‘n’: The product of the first ‘n’ natural numbers is called factorial n and is deonoted by n! i.e, n!=1×2x3x…..x(n-1)xn.
PERMUTATIONEach of the arrangements which can be made by taking some or all of a number of items is called a Permutation.
The permutation of three items a, b,c taken two at a time is ab, ba, bc, cb, ac, ca.
Since the order in which items are taken is important, ab and ba are counted as two different permutations.
The words ‘Permutation’ and ‘Arrangement’ are synonymous and can be used interchangeablyA number of permutation of n things taking r at a time(where n ≤ r) is denoted by nPr and read as nPr.
nPr = \frac{n!}{(n-r)!}COMBINATIONEach of the group or selection which can be made by taking some or all of a number of items is called a Combination.
In combination, the order in which the items are taken is not considered as long as the specific things are included. The combination of three items a, b and c taken two at a time are ab, bc and ca. Here ab and ba are not considered separately because the order in which a and b are taken is not important but
it is only required that combination including a and b is what is to be counted.
The words ‘Combination’ and ‘Selection’ are synonymous.The number of combinations of n dissimilar things taken r at a time is denoted by nCr and read as nCr.
nCr = \frac{n!}{((n-r)!*r!)}.
IMPORTANT RESULTS FOR COMBINATION UNDER DIFFERENT CONDITION1.
nCr =\frac{n!}{r!(n-r)!}2.
nCr = nCn-r3.
nCr + nCr-1 = n+1Cr4. If
nCr = nCs =>
r = s or
n = r + sDISTRIBUTION OF THINGS INTO GROUPS1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is
\frac{(mn)!}{(n!)^m*m!}.
2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is
\frac{(mn)!}{(n!)^m}3. The number of ways in which (m+n+p) things can be divided into three different groups of m,n, and p things respectively is
\frac{(m+n+p)!}{m!n!p!}4. The required number of ways of dividing 3n things into three groups of n each
\frac{(3n)!}{3!*(n!)^3}When the order of groups has importance then the required number of ways =
\frac{(3n)!}{(n!)^3}.
DIVISION OF IDENTICAL OBJECTS INTO GROUPSThe total number of ways of dividing n identical items among r persons, each one of whom, can receive
0, 1, 2 or more items
n+r-1Cr-1The total number of ways of dividing n identical items among r persons, each one of whom receives
at least one item is
n-1Cr-1RESULTS TO REMEBERIn a plane if there are n points of which
no three are collinear, then
1. The number of straight lines that can be formed by joining them is
nC2.
2. The number of triangles that can be formed by joining them is
nC3.
3. The number of polygons with k sides that can be formed by joining them is
nCk.
In a plane if there are n points out of which
m points are collinear, then
1. The number of straight lines that can be formed by joining them is
nC2 - mC2 + 12. The number of triangles that can be formed by joining them is
nC3 - mC33. n straight lines are drawn in the plane such that no two lines are parallel and no three lines three lines are concurrent. Then the number of parts into which these lines divide the plane is equal to
1+\frac{n*(n+1)}{2}nC0 + nC1 + nC2 + nC3 + ..... + nCn = 2^nYou can also check excellent topic by Walker, theory, and links to combinations questions at:
combinations-permutations-and-probability-references-56486.html
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
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