Set S contains nine distinct points in the coordinate plane. If exactl

Given information: Out of 9 points, 5 points lie on the same line.

We can form a triangle using 3 points.

The total number of triangles possible is \(C_3^{9} = 84\).
No triangles can be formed among the 5 points \(C_3^{5} = 10\)

Therefore, the total triangles possible is 74(84 - 10) (Option C)

We have 5 (collinear) points that are on the x-axis and 4 points that are not on the x-axis. Since no set of three points in S is collinear except those on the x-axis, we can have the following 3 cases forming a triangle: 1) two points on the x-axis and one point not on the x-axis, 2) one point on the x-axis and two points not on the x-axis, and 3) three points not on the x-axis.

Case 1: Two points on the x-axis and one point not on the x-axis

There are 5C2 x 4C1 = 10 x 4 = 40 such triangles in this case.

Case 2: one point on the x-axis and two points not on the x-axis

There are 5C1 x 4C2 = 5 x 6 = 30 such triangles in this case.

Case 3: Three points not on the x-axis

There are 4C3 = 4 such triangles in this case.

Therefore, there are 40 + 30 + 4 = 74 triangles that can be formed.

Answer: C
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first consider the scenario with no restrictions..
total points = 9
triangles possible = 9c3 = 84
since five of them lie in x axis,,
the possible combination of 5 points with 3 at a time is 5c3 = 10 (triangles not possible)

total possible traingles = 84 - 10 = 74

Real simply- the wording in this question in saying that 5 points have an x value of zero and therefore line on the x-axis therefore these five points are collinear and cannot form a triangle. Secondly, when this question states no other set of three points in S is collinear- all that really means is that of the 4 points left these four points cannot form a straight line.

Number of Triangles Possible without restriction- also in order to understand the subtraction of the restriction consider this example. Freddie, Shaggy, Velma, Daphne, and Scooby are about to hunt down a ghost but only two members of can be chosen for the hunt. But Freddie and Daphne just got into an argument so they refuse to go together. So the total number of combinations that could be made, and remember order doesn't matter, would be 5c2-2c2=9. When we say order doesn't matter well what that means is Freddie and Scooby in a group represent one possibility- it doesn't matter what position they are in FS or SF because that constitutes the same combination. If we just did 9c3 we would count the total number of possible triangle that include triangles made from the five points.

9c3 - 5c3 = 74

Thus
"C"

*Notice- on the diagram we cannot form a straight line with 3 points that include any of the points (I, II, III, IV)

Answer C

1) There are 4!/(3!1)=4 ways of creating triangle out of the 4 points not on the X axis
2) We know take 2 points not on X axis and 1 on X. 4!/(2!2!)= 6 way of picking 2 points. 5!(1!4!)= 5 ways of picking 1 point from X axis. 6*5=30 ways of picking 1 point from X axis, 2 not from there
3) Now lets count number of ways taking 2 points from X axis and 1 not from there. 5!/(3!2!)= 10 ways to pick 2 points. 4!/(1!3!)= 4 wats to pick 1 point. 4*10=40 triangles

Total number of ways= Sum of all possible triangles =40+30+4=74

In such kind of questions: if there are 'n' total points, out of which 'a' are collinear, the number of triangles formed is always:

nC3 - aC3 , i.e., if all those were collinear then the triangles formed would be nC3, but we have to subtract aC3 to account for those set of points which are not collinear.

So here our answer would be = 9C3 - 5C3 = 84 - 10 = 74.

Thus C is our answer

total ∆ with given points; 9c3 ; 84
and not possible ; 5c3 ; 10
84-10 ; 74
IMO C

Thanks! This is the right method to arrive at the solution.

This is a question on combinations since we need to select 3 points out of a given number to form a triangle. Note that the reordering the points does not give us different triangles, so, we don’t use Permutations concepts here.

The five points lying on the x-axis are collinear. Selecting any 3 points from these will not yield us a triangle. So, this selection set needs to be subtracted from the total number of selections.
The total number of selections of any 3 points taken from 9 points is \(9_C_3\), which is equal to 84. The number of ways of selecting 3 points from the 5 collinear points on the x-axis is \(5_C_3\), which is equal to 10.

Therefore, the total number of triangles = 84 – 10 = 74. The correct answer option is C.

Hope this helps!
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Break up the Possible Triangles into Cases - we must make sure that Each Case does NOT Overlap and that All the Cases are Comprehensive, covering every possible Triangle that can be created.

There are 5 Co-Linear Points on the X-Axis and 4 Points that are in the Coordinate Plane Somewhere that are NOT Co-Linear with any other 2 Points


Case 1: Choose 2 Vertices on the X- Axis ----AND---- Choose 1 Vertex from the 4 points in the Plane

"5 Choose 2" Unique Combinations of way to Choose 2 Points on the X-Axis = 5! / 2!3! = 10

AND

4 Available Options in the Plane to choose for the 3rd Vertex


10 * 4 = 40 Possibilities*****



OR



CASE 2: Pick 1 Vertex on the X-Axis ----- AND------ Pick 2 Vertices from the 4 NON Co-Linear Points in the Plane

"5 Choose 1" = 5 Options

AND

"4 Choose 2" Unique Combinations of 2 Points we can pick in the Plane = 4! / 2!2! = 6 Ways

5 * 6 = 30 Possibilities******


OR


CASE 3: Pick All 3 Points for the Triangle's Vertices from the 4 Points in the Coordinate Plane that are NOT Co-Linear

"4 choose 3" = 4 Possibilities*****




40 + 30 + 4 = 74 Possible Triangles that can be created

-C-

Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?

case 1: Selection only from non linear = 4C3 = 4 ways
case 2: selection from 2 linear and 1 non linear = 5C2*4C1 = 40
case 3: selection from 1 linear and 2 non linear = 5C1*4C2 = 30

Total nos. of ways = 4+40+30 = 74

So, It is C.

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