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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: 58453
Eight dogs are in a pen when a sled owner comes to choose two dogs to  [#permalink]

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20 Dec 2016, 08:00
2
3
00:00

Difficulty:

5% (low)

Question Stats:

82% (01:04) correct 18% (00:56) wrong based on 168 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

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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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07 Jan 2018, 10:29
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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Joined: 15 Nov 2017
Posts: 52
Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to  [#permalink]

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09 Dec 2018, 16:03
Hi Bunuel,

Is there a way to tell when a problem is going to be a C vs. a P? For instance, it seems in this problem that the words "different ordering" lead to the conclusion of C; however, sometimes I have trouble confidently determining this.

Your insight would be great. Thanks!

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
Senior Manager
Joined: 17 Mar 2014
Posts: 429
Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to  [#permalink]

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08 Jan 2019, 20:25
KHow wrote:
Hi Bunuel,

Is there a way to tell when a problem is going to be a C vs. a P? For instance, it seems in this problem that the words "different ordering" lead to the conclusion of C; however, sometimes I have trouble confidently determining this.

Your insight would be great. Thanks!

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

KHow
whenever ordering matters it is P and if ordering doesn't matter then it is C.

for example, in this question, it is explicitly mentioned that "different orderings of the same dogs are considered the same team" it means ordering doesn't matter.

Tom - Jerry
Jerry-Tom

Both are same team as per above question so it is C.
Re: Eight dogs are in a pen when a sled owner comes to choose two dogs to   [#permalink] 08 Jan 2019, 20:25
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