In how many ways can $5$ boys and $5$ girls stand in a row so that no two girls maybe together?

The correct option is C

$5!\times 6!$

Explanation for the correct option:

Finding the required number of ways:

The number of boys$=5$

The number of girls $=5$

The number of ways boys can be arranged $=5!$

Let ‘B' denotes the position of boys and ‘G’ denotes the position of girls.

Then the arrangement of girls standing in a row where no two girls may be together is as follows

_ B G B G B G B G B G

There are $6$ positions for girls where no two girls are together

So, the number of ways, the girls can be arranged $=6!$

Hence the required number of ways $=5!\times 6!$

Thus, option (C) is correct.