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# 10 Arabian horses are split into pairs to pull one of the

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Manager
Joined: 25 Nov 2009
Posts: 57
Location: India
10 Arabian horses are split into pairs to pull one of the  [#permalink]

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Updated on: 30 Jun 2013, 12:40
1
1
00:00

Difficulty:

75% (hard)

Question Stats:

48% (02:37) correct 52% (02:39) wrong based on 51 sessions

### HideShow timer Statistics

10 Arabian horses are split into pairs to pull one of the distinct 4 carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible?

A, 420
B. 1260
C. 5220
D. 9450
E. 113400

Note: Found this question. But not the OA. Can anyone help? Thanks in advance.

Originally posted by jusjmkol740 on 06 Jun 2010, 07:57.
Last edited by Bunuel on 30 Jun 2013, 12:40, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 51101
Re: Pairs of Horses and Carts  [#permalink]

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06 Jun 2010, 08:37
2
jusjmkol740 wrote:
10 Arabian horses are split into pairs to pull one of the distinct 4 carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible?

a) 420
b) 1260
c) 5220
d) 9450
e) 113400

Note: Found this question. But not the OA. Can anyone help? Thanks in advance.

Note: The question is beyond the GMAT scope.

# of ways 10 horses can be divided into 5 groups when order of the groups does not matter is: $$\frac{C^2_{10}*C^2_{8}*C^2_{6}*C^2_{4}*C^2_{2}}{5!}=945$$.

So we would have 945 different cases of dividing 10 horses into 5 groups:
1. $$A_1$$, $$B_1$$, $$C_1$$, $$D_1$$, $$E_1$$ (each letter represent pair of horses);
2. $$A_2$$, $$B_2$$, $$C_2$$, $$D_2$$, $$E_2$$;
...
945. $$A_{945}$$, $$B_{945}$$, $$C_{945}$$, $$D_{945}$$, $$E_{945}$$.

Now, we want to assign a pair (a letter in our case) to one of 4 distinct carts. For each case (out of 945) $$P^4_5=120$$ would represent # of ways to choose 4 pairs out of 5 when order matters (as carts are distinct). OR $$C^4_5$$ - choose 4 different letters out of 5 and $$4!$$ - # of ways to assign 4 different letters to 4 distinct carts: $$C^4_5*4!=120$$.

So total # of different assignments would be $$945*120=113400$$.

probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
probability-88685.html#p669025
combination-groups-and-that-stuff-85707.html#p642634
sub-committee-86346.html?highlight=divide+groups
combinations-problems-95344.html#p733805

Hope it helps.
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Posts: 9121
Re: 10 Arabian horses are split into pairs to pull one of the  [#permalink]

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12 Dec 2017, 23:34
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Re: 10 Arabian horses are split into pairs to pull one of the &nbs [#permalink] 12 Dec 2017, 23:34
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