BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Counting

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

Counting

by knight247 » Mon Sep 26, 2011 11:26 pm
Ten Horses are split into pairs to pull one of the distinct four carts in a race. If each cart is assigned to a pair of horses, how many different assignments of horses to carts are possible?
(A)420
(B)1260
(C)5220
(D)9450
(E)113400

OA is E

I first arranged the horses as H1 H2 H3 H4 H5 H6 H7 H8 H9 H10
Then the carts as................. A...A...B...B...C...C...D...D...R...R

Where R stands for rejected

Now the carts can be arranged as
10!/(2!2!2!2!2!)=113400 Hence E

I'm hoping someone can help me with an alternate method for this one. Something quicker. Detailed explanations would be appreciated. Thanks
Top
Source: — Problem Solving |
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Mon Sep 26, 2011 11:31 pm
Total 10 horses, need to pick 2 from 10 for a cart;

You have;

10C2 * 8C2 * 6C2 * 4C2 = 113400
Top
Legendary Member
Posts: 627
Joined: Thu Jun 23, 2011 9:12 am
Thanked: 4 times
Followed by:1 members

by mankey » Tue Sep 27, 2011 10:03 am
This one is not clear. Some expert please help.

Thanks
Mankey
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 312
Joined: Tue Aug 02, 2011 3:16 pm
Location: New York City
Thanked: 130 times
Followed by:33 members
GMAT Score:780

by gmatboost » Tue Sep 27, 2011 9:32 pm
Ten Horses are split into pairs to pull one of the distinct four carts in a race. If each cart is assigned to a pair of horses, how many different assignments of horses to carts are possible?
Shankar's approach is the way I approached it as well.

When we choose a pair of horses to pull each cart, since order does not matter, it is a combination (rather than a permutation).

For the first cart, we have 10 horses to choose from, and we must choose 2. This is 10C2 = 10!/(8!2!) = 10*9/2 = 45.

For the second cart, we have only 8 horses left to choose from, and we must choose 2. This is 8C2 = 8!/(6!2!) = 8*7/2 = 28.

For the third cart, we have only 6 horses left to choose from, and we must choose 2. This is 6C2 = 6!/(4!2!) = 6*5/2 = 15.

And, for the fourth and final cart, we have only 4 horses left to choose from, and we must choose 2. This is 4C2 = 4!/(2!2!) = 4*3/2 = 12.

We do not need to do anything further for the last two horses. This covers all possible pairs of horses for each of the carts.

45*28*15*6 is an annoying thing to calculate.
I recommend pairing the math as follows:
45*6 = 270.
28*15 = 420.
270*420 = 113400.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
Top
Post new topic Post Reply
• Page 1 of 1