Question

What are the number of ways of making 5 couples stand in a line such that no man is positionedahead of his wife?​

Answer 1

Answer:

woman stands one side and man stand another side

Answer 2

Given,

The number of couples$\[ = 5\]$

Solution,

Understand that there are $\[10\]$ people and they can stand-in $\[10!.\]$

Know that for any two positions out of the ten occupied by a man and his wife, the pair cannot rearrange amongst themselves in $\[2!\]$ and there are five couples.

Therefore,

$\[ \Rightarrow \frac{{10!}}{{2! \times 2! \times 2! \times 2! \times 2!}}\]$

Hence, the number of ways are $\[\frac{{10!}}{{2! \times 2! \times 2! \times 2! \times 2!}}.\]$