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

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by swerve » Wed Dec 06, 2017 10:19 am
Eight dogs are in a pen when a sled owner comes to choose six 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. 720
B. 120
C. 56
D. 28
E. 6

The OA is D.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
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by [email protected] » Wed Dec 06, 2017 10:23 am
Hi swerve,

We're told to choose 6 dogs from a group of 8 dogs to form a sled team and different orderings of the same dogs are considered the SAME team. We're asked for the number of different sled teams that can be formed. Since the 'order' of the dogs does NOT matter, this is a Combination Formula question:

8!/6!2! = (8)(7)/(2)(1) = 56/2 = 28 possible teams

Final Answer: D

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by Scott@TargetTestPrep » Sun Sep 29, 2019 6:06 pm
swerve wrote:Eight dogs are in a pen when a sled owner comes to choose six 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. 720
B. 120
C. 56
D. 28
E. 6

The OA is D.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
We are given that there are 8 dogs and the owner needs to arrange 6 of them. Since different orderings of the same dogs are considered the same team order does not matterand we have a combination problem. Thus, the number of ways to choose 6 dogs from 8 is:

8C6 = 8!/[(8 - 6)!*6!] = (8 x 7)/2! = 4 x 7 = 28

Answer: D

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