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

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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.

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

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