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Eight dogs are in a pen when a sled owner comes to choose two dogs to

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Eight dogs are in a pen when a sled owner comes to choose two dogs to [#permalink]

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Eight dogs are in a pen when a sled owner comes to choose two dogs to form a sled team. If the dogs are to be placed in a straight line and different orderings of the same dogs are considered the same team, how many different sled teams can the owner form?

A. 72
B. 56
C. 42
D. 30
E. 28
[Reveal] Spoiler: OA

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Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to [#permalink]

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New post 07 Jan 2018, 09:29
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose two dogs to form a sled team. If the dogs are to be placed in a straight line and different orderings of the same dogs are considered the same team, how many different sled teams can the owner form?

A. 72
B. 56
C. 42
D. 30
E. 28


Since there are eight dogs in a pen and we need to choose 2 of these dogs to form a sled team,

there are \(C^2_8 = \frac{8!}{6!*2!} = \frac{8*7}{2} = 28\) ways to choose the teams(Option E)
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Kudos [?]: 830 [0], given: 22

Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to   [#permalink] 07 Jan 2018, 09:29
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Eight dogs are in a pen when a sled owner comes to choose two dogs to

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