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

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

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06 Jun 2010, 07:57
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55% (03:04) correct 45% (01:37) wrong based on 37 sessions

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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jun 2013, 12:40, edited 1 time in total.
Added the OA
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Re: Pairs of Horses and Carts [#permalink]

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06 Jun 2010, 08:37
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Expert's post
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 $$945*120=113400$$.

Answer: E.

Questions about the same concept:

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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Re: Pairs of Horses and Carts [#permalink]

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07 Jun 2010, 11:42
Thanks Bunnel.
But I'm still a bit confused about the explanation. Some more elaboration would certainly help.
Math Expert
Joined: 02 Sep 2009
Posts: 37082
Followers: 7242

Kudos [?]: 96315 [0], given: 10737

Re: Pairs of Horses and Carts [#permalink]

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07 Jun 2010, 11:49
jusjmkol740 wrote:
Thanks Bunnel.
But I'm still a bit confused about the explanation. Some more elaboration would certainly help.

Please go through the links in my previous post to understand the concept, also see the newest topic with similar problem at: combinations-problems-95344.html.

Hope it helps.
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Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
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Ten Arabian horses are split into pairs to pull one of the distinct [#permalink]

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12 Sep 2014, 10:33
Ten 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 of horses, how many different assignments of
horses to cart are possible?

a 420
b 1260
c 5220
d 9450
e 113400
Math Expert
Joined: 02 Sep 2009
Posts: 37082
Followers: 7242

Kudos [?]: 96315 [0], given: 10737

Re: 10 Arabian horses are split into pairs to pull one of the [#permalink]

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12 Sep 2014, 10:49
manish2014 wrote:
Ten 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 of horses, how many different assignments of
horses to cart are possible?

a 420
b 1260
c 5220
d 9450
e 113400

Merging topics. Please do NOT post questions from questionable sources.
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Re: 10 Arabian horses are split into pairs to pull one of the   [#permalink] 12 Sep 2014, 10:49
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# 10 Arabian horses are split into pairs to pull one of the

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