# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
 ✖ A. $\dfrac{6!}{3! × 3! × 3!}$ ✖ B. $\dfrac{6!}{2! × 2!}$ ✔ C. $\dfrac{6!}{2! × 2! × 2!}$ ✖ D. $\dfrac{6!}{2! × 3!}$

Solution:
Option(C) is correct

6 people can be made to stand in a line in $6!$ Ways.

However, the problem introduces a constraint that no man stands in a position that is ahead of his wife.

For any 2 given positions out of the 6 occupied by a man and his wife, the pair cannot re-arrange amongst themselves in $2!$ Ways as the wife has to be in a position ahead of the man. Only one of the $2!$ arrangements is allowed.

As there are 3 couples in the group, the total number of ways gets reduced by a factor of $(2!×2!×2!)$.
Hence, the total number of ways,

$= \dfrac{6!}{2! × 2! × 2!}$

## (1) Comment(s)

Shubham Kumar
()

Does anybody have an easier explanation to the question apart from the given one?