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Joined: 02 Sep 2009
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15 Sep 2014, 23:20
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Difficulty:

35% (medium)

Question Stats:

69% (01:01) correct 31% (01:21) wrong based on 189 sessions

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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

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15 Sep 2014, 23:20
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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).

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01 Jun 2015, 05:22
D

C315−(C38+C37)=455−(56+35)=364
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Apoorv

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13 Dec 2015, 17:13
C28∗C17=28∗7=196 What formula is the first C (which equals 28)?
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Joined: 02 Sep 2009
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17 Dec 2015, 08:26
LostinNY wrote:
C28∗C17=28∗7=196 What formula is the first C (which equals 28)?

C stands for combinations. Check for more here: math-combinatorics-87345.html
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10 Jan 2016, 18:46
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.
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18 May 2016, 13:47
Math Expert
Joined: 02 Aug 2009
Posts: 7106

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18 May 2016, 19:48
1
acarpio1 wrote:

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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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

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Location: Spain
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15 Jan 2018, 10:05
This post is just to say that this question is categorized as 600 level in the Quizzs.
Thank you
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Joined: 08 Aug 2018
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GMAT 1: 210 Q1 V1

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02 Sep 2018, 08:13
Hard for a 600 Level
Re: M03-23 &nbs [#permalink] 02 Sep 2018, 08:13
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