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

Question

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

GMATinsight · Accepted Answer

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

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

ShreyasJavahar · Answer

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?

Gaurav2896 · Answer

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!.

bumpbot · Answer

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

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