Question

Find the number of ways in which $5$ boys and $5$ girls may be seated in a row so that no two girls and no two boys are together.

Answer 1

Find the number of ways as per the given conditions

From the given data, there are $5$ boys and $5$ girls.

The permutation is an arrangement of $\mathrm{r}$ items taken out of $\mathrm{n}$ items and the combination is the selection of $\mathrm{r}$ items taken out of $\mathrm{n}$ items irrespective of the arrangement.

The permutation formula is,

${}^{\mathrm{n}}\mathrm{P}_{\mathrm{r}}=\frac{\mathrm{n}!}{\left(\mathrm{n}-\mathrm{r}\right)!}$

So, the number of boys seated in a row in,

${}^5\mathrm{P}_5=5!$ ways

The girls can fill the $6$ space in ${}^6\mathrm{P}_5=6!$ ways.

So,

The number of ways in which no $2$ girls and no $2$ boys sit together $=5!\times 6!$

Hence, the number of ways in which $5$ boys and $5$ girls may be seated in a row so that no two girls and no two boys are together is $5!\times 6!$.