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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] New post   01 Apr 2020, 23:40
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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] New post   02 Apr 2020, 00:48
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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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There are 4 men, 3 women and 3 children seated at a round picnic table [#permalink] New post   02 Oct 2021, 12:03
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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


HI your answer is correct but you reasoning when you do Remaining 7 people can sit in 7! ways on remaining empty chairs is incorrect .
In circular no of ways is (n-1)! . Here 7 people will remain and one group of 3 people (containing 1Man , 1Female and 1Child) (after arranging 3 ).
So total ways of arranging these 8 in circular will be (8-1)! = 7!
So total ways is 24*7!.
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Re: There are 4 men, 3 women and 3 children seated at a round picnic table [#permalink] New post   04 Apr 2020, 02:42
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] New post   22 Oct 2023, 00:39
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Re: There are 4 men, 3 women and 3 children seated at a round picnic table [#permalink]
22 Oct 2023, 00:39
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