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Permutation - Find the number of ways in which four men, two women and

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Intern
Joined: 06 Aug 2018
Posts: 8
Permutation - Find the number of ways in which four men, two women and  [#permalink]

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26 Aug 2018, 07:23
Hello All,

Question: Find the number of ways in which four men, two women and a child can stand in a row if the child is standing between the two women?

My solution:
Child can sit in 3 locations that will allow them to have two spaces beside him. Once the child is seated the first woman has two choices and the second woman has one choice. Once the women are seated there remains 4 seats and 4 men.
This gives a total possible arrangements of = 3*2!*4! = 144 ways

In case 2, you have to arrange 4 men and a group which you can do in 5! ways. The group consists of two women and a child between them. The two women can be arranged around the child in 2 ways. So total ways = 2*5! = 240 ways

I would like to know where I went wrong

Source of question (case 2): https://gmatclub.com/forum/combinatoric ... 06266.html
Source of answer: https://www.veritasprep.com/blog/2011/1 ... 3-part-ii/
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Intern
Joined: 06 Aug 2018
Posts: 8
Re: Permutation - Find the number of ways in which four men, two women and  [#permalink]

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26 Aug 2018, 08:08
I think I spotted my error the child can sit in 5 locations not 3 locations as mentioned above. This leads us to the correct solution:

= 5*2!*4!
= 240 ways
Re: Permutation - Find the number of ways in which four men, two women and   [#permalink] 26 Aug 2018, 08:08
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