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?

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