Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 06:41 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M03-23

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56251

### Show Tags

3
19 00:00

Difficulty:   35% (medium)

Question Stats: 69% (01:15) correct 31% (01:33) wrong based on 265 sessions

### HideShow timer Statistics 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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56251

### Show Tags

3
5
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).

_________________
Manager  Joined: 21 May 2015
Posts: 221
Concentration: Operations, Strategy
GMAT 1: 750 Q50 V41 ### Show Tags

D

C315−(C38+C37)=455−(56+35)=364
_________________
Apoorv

I realize that i cannot change the world....But i can play a part Intern  Joined: 13 Nov 2011
Posts: 18

### Show Tags

C28∗C17=28∗7=196 What formula is the first C (which equals 28)?
Math Expert V
Joined: 02 Sep 2009
Posts: 56251

### Show Tags

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
_________________
Manager  Joined: 01 Aug 2014
Posts: 54
Schools: Rotman '17 (A)
GMAT 1: 710 Q44 V42 ### Show Tags

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.
Intern  Joined: 06 Feb 2016
Posts: 14

### Show Tags

Math Expert V
Joined: 02 Aug 2009
Posts: 7764

### Show Tags

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
_________________
Intern  B
Joined: 24 Oct 2014
Posts: 14
Location: Spain
Schools: LBS '21 (II)

### Show Tags

This post is just to say that this question is categorized as 600 level in the Quizzs.
Thank you
Intern  B
Joined: 08 Aug 2018
Posts: 6
GMAT 1: 210 Q1 V1 ### Show Tags

Hard for a 600 Level Re: M03-23   [#permalink] 02 Sep 2018, 09:13
Display posts from previous: Sort by

# M03-23

Moderators: chetan2u, Bunuel  