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
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Hard for a 600 Level