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There are 4 men, 3 women and 3 children seated at a round picnic table

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There are 4 men, 3 women and 3 children seated at a round picnic table  [#permalink]

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New post 01 Apr 2020, 22:40
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  55% (hard)

Question Stats:

50% (02:21) correct 50% (02:29) wrong based on 16 sessions

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There are 4 men, 3 women and 3 children seated at a round picnic table. How many ways can a specific child be seated between a man and a woman?

A) \(7! × 24\)
B) \(8! × 12\)
C) \(\frac{9!}{(2 × (4! × 3!))}\)
D) \(10! - (3 × 3C2) - (3 × 4C2)\)
E) \(3!^3\)
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Re: There are 4 men, 3 women and 3 children seated at a round picnic table  [#permalink]

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New post 01 Apr 2020, 23:48
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costcosized wrote:
There are 4 men, 3 women and 3 children seated at a round picnic table. How many ways can a specific child be seated between a man and a woman?

A) \(7! × 24\)
B) \(8! × 12\)
C) \(\frac{9!}{(2 × (4! × 3!))}\)
D) \(10! - (3 × 3C2) - (3 × 4C2)\)
E) \(3!^3\)


One specific child is to be fixed

Now, we need to pick a man and a woman to sit adjacent to the selected child - 4C1* 3C1

The chosen man and woman can sit in 2! ways adjacent to the d=child [either man left and woman right or vice versa]

Remaining 7 people can sit in 7! ways on remaining empty chairs

Total required ways of seating arrangements = 4C1* 3C1*2!*7! = 24*7!

Answer: Option A
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Re: There are 4 men, 3 women and 3 children seated at a round picnic table  [#permalink]

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New post 04 Apr 2020, 01:42
GMATinsight wrote:
costcosized wrote:
There are 4 men, 3 women and 3 children seated at a round picnic table. How many ways can a specific child be seated between a man and a woman?

A) \(7! × 24\)
B) \(8! × 12\)
C) \(\frac{9!}{(2 × (4! × 3!))}\)
D) \(10! - (3 × 3C2) - (3 × 4C2)\)
E) \(3!^3\)


One specific child is to be fixed

Now, we need to pick a man and a woman to sit adjacent to the selected child - 4C1* 3C1

The chosen man and woman can sit in 2! ways adjacent to the d=child [either man left and woman right or vice versa]

Remaining 7 people can sit in 7! ways on remaining empty chairs

Total required ways of seating arrangements = 4C1* 3C1*2!*7! = 24*7!

Answer: Option A


Hello,
Don't you have to take into account the 3 ways in which a child can be selected?
I agree that in this question, that is not even in the answer choices, but say we had just plain old integers in the choices and the integer equivalents of 24*7! and 24*3*7! were amongst the choices, how would one go about determining whether the 3 ways in which a child can be chosen should be accounted for or not?
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Re: There are 4 men, 3 women and 3 children seated at a round picnic table   [#permalink] 04 Apr 2020, 01:42

There are 4 men, 3 women and 3 children seated at a round picnic table

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