Thirteen distinct points lie in a plane with exactly n of the points

Question

Thirteen distinct points lie in a plane with exactly n of the points lying on the same line and the other 13-n points non-collinear. If 276 unique triangles can be drawn using these points as vertices, what is the value of n?

A. 2
B. 3
C. 4
D. 5
E. 7

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A. 2
B. 3
C. 4
D. 5
E. 7

ScottTargetTestPrep · Accepted Answer

If all 13 points are non-collinear, then 13C3 = (13 x 12 x 11) / (3 x 2) = 13 x 2 x 11 = 286 unique triangles can be drawn. However, since only 276 triangles can be drawn, 10 triangles must have been created by the n collinear points. Therefore, we can create the equation:

nC3 = 10

n! /

n(n - 1)(n - 2) / (3 x 2) = 10

n(n - 1)(n - 2) = 60

From the above equation, we see n must be 5 since 5(4)(3) = 60.

Answer: D

Archit3110 · Answer

total ∆ ; 13c3 ; 286
and since 276 unique ∆ are to be formed ; 286-nc3 ; 276
n has to be 5 
IMO D

stne · Answer

n_{c_1} *{(13-n)}_{c_2} +n_{c_2} *{(13-n)}_{c_1}+{(13-n)}_{c_3}  =276

Where  n_{c_1}  = Choose 1 point from the n collinear points 
 {(13-n)}_{c_2} = Choose 2 points from the (13-n ) NON-collinear points 
 {(13-n)}_{c_3}  = Choose 3 points from the (13-n) NON-collinear points

It is better to substitute answer choices after this , we see that for n = 5 , the above equation gives 276

Ans-D

Hope it's clear.

Kinshook · Answer

Thirteen distinct points lie in a plane with exactly n of the points lying on the same line and the other 13-n points non-collinear.

If 276 unique triangles can be drawn using these points as vertices, what is the value of n?

Total triangles = 13C3
Number of triangles not feasible since all 3 points are collinear = nC3
Total feasible triangles = 13C3 - nC3 = 276

286 - n(n-1)(n-2)/6 = 276
n(n-1)(n-2) = 60 
n = 5; n(n-1)(n-2) = 5*4*3 = 60

IMO D

Thirteen distinct points lie in a plane with exactly n of the points

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Thirteen distinct points lie in a plane with exactly n of the points

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