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# Eight dogs are in a pen when the owner comes to walk some of them. The

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Math Expert
Joined: 02 Sep 2009
Posts: 46264
Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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26 Apr 2017, 09:08
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Difficulty:

15% (low)

Question Stats:

76% (00:37) correct 24% (01:01) wrong based on 69 sessions

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Eight dogs are in a pen when the owner comes to walk some of them. The owner lets six dogs out of the pen one at a time. How many different variations in the group of dogs he takes walking are possible?

A. 28
B. 56
C. 560
D. 1,680
E. 6,720

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Joined: 22 Jun 2016
Posts: 250
Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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26 Apr 2017, 10:35
Similar question posted few days back.

https://gmatclub.com/forum/eight-dogs-a ... 37455.html
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Intern
Joined: 14 Mar 2017
Posts: 5
Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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18 May 2017, 03:00
I think the OA should be E 8*7*6*5*4*3 ? Where am I missing something ?
Math Expert
Joined: 02 Sep 2009
Posts: 46264
Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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18 May 2017, 03:26
ppcm wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets six dogs out of the pen one at a time. How many different variations in the group of dogs he takes walking are possible?

A. 28
B. 56
C. 560
D. 1,680
E. 6,720

I think the OA should be E 8*7*6*5*4*3 ? Where am I missing something ?

Notice that we are asked about the number of groups, so we are not interested in the different arrangements within each group. Thus, the answer is simply $$C^6_8=\frac{8!}{(8-6)!*6!}=28$$.

Moreover, even if it were how many arrangements of 6 dogs are possible (for example the way we have HERE), the answer then would be $$C^6_8*6!=\frac{8!}{2!6!}*6!=28*6!$$ or simply $$P^6_8=\frac{8!}{(8-6)!}$$.

Hope it's clear.
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Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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18 May 2017, 03:51
Bunuel wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets six dogs out of the pen one at a time. How many different variations in the group of dogs he takes walking are possible?

A. 28
B. 56
C. 560
D. 1,680
E. 6,720

Since we are talking about different variations in the group of dogs ..
So such combination can be formed in $$8C6$$ = 8*7/2 = 28 ways..

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Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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18 May 2017, 09:40
Bunuel wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets six dogs out of the pen one at a time. How many different variations in the group of dogs he takes walking are possible?

A. 28
B. 56
C. 560
D. 1,680
E. 6,720

$$8C_6 = 28$$

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Re: Eight dogs are in a pen when the owner comes to walk some of them. The [#permalink]

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22 May 2017, 18:33
Bunuel wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets six dogs out of the pen one at a time. How many different variations in the group of dogs he takes walking are possible?

A. 28
B. 56
C. 560
D. 1,680
E. 6,720

We need to determine how many ways the owner can select 6 dogs from 8. The order within the group of the 6 dogs he selects doesn’t matter, so the number of ways he can select 6 dogs from 8 is 8C6:

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

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Re: Eight dogs are in a pen when the owner comes to walk some of them. The   [#permalink] 22 May 2017, 18:33
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