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

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Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139709 [2], given: 12794

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

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20 Dec 2016, 07:00
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
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Difficulty:

15% (low)

Question Stats:

77% (00:59) correct 23% (00:41) wrong based on 65 sessions

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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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Kudos [?]: 139709 [2], given: 12794

BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 1815

Kudos [?]: 830 [0], given: 22

Location: India
WE: Sales (Retail)
Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to [#permalink]

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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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Stay hungry, Stay foolish

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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