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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: 51258
Eight dogs are in a pen when the owner comes to walk some of them. The  [#permalink]

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02 Apr 2017, 23:33
00:00

Difficulty:

15% (low)

Question Stats:

80% (01:13) correct 20% (00:56) 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 four dogs out of the pen one at a time. How many different variations in the line of dogs leaving the pen are possible?

A. 3,360
B. 1,680
C. 540
D. 60
E. 24

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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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03 Apr 2017, 02:36
IMO B
You need to use permutation in this case
8!/(8-4)!=1680
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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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06 Apr 2017, 08:33
Bunuel wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets four dogs out of the pen one at a time. How many different variations in the line of dogs leaving the pen are possible?

A. 3,360
B. 1,680
C. 540
D. 60
E. 24

We are given that there are 8 dogs and the owner needs to arrange 4 of them. Since we have an “order arrangement,” order matters. So we have a permutation problem. Thus, the number of ways to arrange 4 dogs from 8 is:

8P4 = 8!/(8-4)! = 8 x 7 x 6 x 5 = 1,680

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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, 02:55
Bunuel wrote:
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets four dogs out of the pen one at a time. How many different variations in the line of dogs leaving the pen are possible?

A. 3,360
B. 1,680
C. 540
D. 60
E. 24

First we should make a group of 4 dogs which will be let out. Then we will find how many arrangements of such leaving will be possible.

So no. of ways in which a group of 4 dogs can be formed = $$8C4$$ = (8*7*6*5)/(4*3*2*1) = 70

The total no. of ways in which the line of dogs leaving the pen are possible = 70 X 4! = 70*4*3*2= 1680

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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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26 Nov 2018, 11:33
No of ways to arrange 4 from 8
8P4 = 1680 (B)
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Re: Eight dogs are in a pen when the owner comes to walk some of them. The &nbs [#permalink] 26 Nov 2018, 11:33
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