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Jack Danison, his wife and his two kids went to a theatre to watch a

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Math Expert
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Jack Danison, his wife and his two kids went to a theatre to watch a  [#permalink]

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New post 25 Jun 2019, 03:02
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

64% (00:46) correct 36% (00:48) wrong based on 56 sessions

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Joined: 22 Nov 2018
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Re: Jack Danison, his wife and his two kids went to a theatre to watch a  [#permalink]

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New post 25 Jun 2019, 03:06
1
Bunuel wrote:
Jack Danison, his wife and his two kids went to a theatre to watch a movie. In how many ways can they sit in a row such that the two kids have to be together always?

A. 3
B. 6
C. 9
D. 12
E. 18


Consider the 2 kids as one then 3 people can be seated in 3! ways
2 kids can be seated among themselves in 2 ways so total is 3!*2=12 IMO D
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Re: Jack Danison, his wife and his two kids went to a theatre to watch a  [#permalink]

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New post 25 Jun 2019, 05:45
Bunuel wrote:
Jack Danison, his wife and his two kids went to a theatre to watch a movie. In how many ways can they sit in a row such that the two kids have to be together always?

A. 3
B. 6
C. 9
D. 12
E. 18


Number of ways of arranging M, W, [\(K_1\), \(K_2\)] = 3!*2! = 6*2 = 12 ways

IMO Option D
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Re: Jack Danison, his wife and his two kids went to a theatre to watch a  [#permalink]

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New post 26 Jun 2019, 03:09
Bunuel wrote:
Jack Danison, his wife and his two kids went to a theatre to watch a movie. In how many ways can they sit in a row such that the two kids have to be together always?

A. 3
B. 6
C. 9
D. 12
E. 18


total ways for kids together consider them as 1 ; KKJW ; 3! *2 ways ;
IMO D ; 12
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Re: Jack Danison, his wife and his two kids went to a theatre to watch a  [#permalink]

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New post 01 Jul 2019, 18:04
2
Bunuel wrote:
Jack Danison, his wife and his two kids went to a theatre to watch a movie. In how many ways can they sit in a row such that the two kids have to be together always?

A. 3
B. 6
C. 9
D. 12
E. 18


We need to determine how many ways we can arrange:

[K1 - K2] - [J] - [W]

We see that we have 3 total slots, which can be arranged in 3! = 6 ways, and the two kids can be arranged in 2! = 2 ways. Thus, the total number of arrangements is 2 x 6 = 12.

Answer: D
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Re: Jack Danison, his wife and his two kids went to a theatre to watch a   [#permalink] 01 Jul 2019, 18:04
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