# 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. $$\frac{{6!}}{{3! \times 3! \times 3!}}$$

B. $$\frac{{6!}}{{2! \times 2!}}$$

C. $$\frac{{6!}}{{2! \times 2! \times 2!}}$$

D. $$\frac{{6!}}{{2! \times 3!}}$$

**Answer: Option C **

__Solution(By Examveda Team)__

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 rearrange 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,

= $$\frac{{6!}}{{2! \times 2! \times 2!}}$$

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