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?