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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

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

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Source: Beat The GMAT — Problem Solving | Mon Sep 26, 2011 11:26 pm

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

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

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mankey: Legendary Member; Posts: 627; Joined: Thu Jun 23, 2011 9:12 amFollowed by: 1 member; Thanked: 4 times

by mankey » Tue Sep 27, 2011 10:03 am

This one is not clear. Some expert please help.

Thanks
Mankey

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gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; 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.

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