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A canyon trail ride company owns 4 horses and 8 mules. If every group

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A canyon trail ride company owns 4 horses and 8 mules. If every group [#permalink]

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A canyon trail ride company owns 4 horses and 8 mules. If every group of animals they send on a ride must contain exactly 2 horses and 3 mules, how many different groups of animals could they send?

A) 32
B) 56
C) 72
D) 192
E) 336
[Reveal] Spoiler: OA

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Re: A canyon trail ride company owns 4 horses and 8 mules. If every group [#permalink]

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New post 18 Oct 2017, 12:56
4C2*8C3 = 6*56 = 336 .

Hence answer is E.
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Re: A canyon trail ride company owns 4 horses and 8 mules. If every group [#permalink]

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New post 18 Oct 2017, 17:25
4C2*8C3 = 6*56 = 336 .

Hence answer is E.

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Re: A canyon trail ride company owns 4 horses and 8 mules. If every group [#permalink]

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souvonik2k wrote:
A canyon trail ride company owns 4 horses and 8 mules. If every group of animals they send on a ride must contain exactly 2 horses and 3 mules, how many different groups of animals could they send?

A) 32
B) 56
C) 72
D) 192
E) 336


The number of ways to select 3 mules from 8 mules is 8C3 = 8!/[3!(8-3)!] = (8 x 7 x 6)/3! = 56.

The number of ways to select 2 horses from 4 hours is 4C2 = 4!/[2!(4-2)!] = (4 x 3)/2! = 6.

Thus, the number of ways to select a group consisting of 3 mules and horses is 56 x 6 = 336.

Answer: E
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Re: A canyon trail ride company owns 4 horses and 8 mules. If every group [#permalink]

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New post 28 Oct 2017, 23:02
The number of ways to select 3 mules from 8 mules is 8C3 = 8!/[3!(8-3)!] = (8 x 7 x 6)/3! = 56.

The number of ways to select 2 horses from 4 hours is 4C2 = 4!/[2!(4-2)!] = (4 x 3)/2! = 6.

Thus, the number of ways to select a group consisting of 3 mules and horses is 56 x 6 = 336.

Answer: E

Kudos [?]: 7 [0], given: 186

Re: A canyon trail ride company owns 4 horses and 8 mules. If every group   [#permalink] 28 Oct 2017, 23:02
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