Official Solution:

If one of two parallel lines has 7 points on it and the other has 8, how many unique triangles can be drawn using the points?

A. 168
B. 196
C. 316
D. 364
E. 455


Approach #1:

There are two types of triangles possible:

With two vertices on the line with 8 points and the third vertex on the line with 7 points: \(C^2_8*C^1_7=28*7=196\);

With two vertices on the line with 7 points and the third vertex on the line with 8 points: \(C^2_7*C^1_8=21*8=168\);

Total: \(196+168=364\).

Approach #2:

All different 3 points out of total \(8+7=15\) points will create a triangle EXCEPT those 3 points which are collinear.

\(C^3_{15}-(C^3_8+C^3_7)=455-(56+35)=364\) (where \(C^3_8\) and \(C^3_7\) are # of different 3 collinear points possible from the line with 8 points and the line with 7 points, respectively).


Answer: D
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C28∗C17=28∗7=196 What formula is the first C (which equals 28)?

I am very familiar with the combinatoric method. My approach to this problem was (7*8!/2! + 8*&7!/2!) but this was clearly wrong.

I have read the combinatorics section of the gmatclub site but still don't understand how to derive the proper combinatoric equation from the C with the two small numbers that is in the answer explanation above.

Daft, I know... Thanks for any help.

Hi acarpio1 and Anonamy,

The Q talks of two parallel lines with 7points and 8 points on each...
Now we have to form triangles..

We cannot form a triangle from all three points on the same line..
so we have to take one from 1 line and 2 from second line = 7C1*8C2..
and in second case 2 from ONE and 1 from SECOND = 7C2*8C1...
so 7C1 *8C2 +8C1 * 7C2 = 7*7*8/2 + 8*7*6/2 = 49*4 + 8*21 = 364
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Hard for a 600 Level