Question

In how many ways can $5$ boys and $5$ girls sit in a circle so that no two boys sit together?
$A.\;5!\; \times \;5! \\\ B.\;4!\; \times \;5! \\\ C.\;\dfrac{{5!\; \times \;5!}}{2} \\\$
$D.$ None of these

Answer 1

In this we apply the concept of circular permutation i.e. the number of circular permutations of $m$ things of first kind , $n$ things of second kind, such that no two things of second kind come together is $\left( {m - 1} \right)!\; \times \;{}^m{P_n}$ ways.

Complete step-by-step answer:

According to the question we have to find out how many ways can $5$ boys and $5$ girls can sit in a circle so that no two boys sit together.

Hence , what we have to do first is to fix the alternate position of girls as shown in the figure .

Five girls can be seated around the circle in $\left( {5 - 1} \right)! = 4!$

Five boys can be seated in five – vacant places for $5!$.

$\therefore$ Required number of ways $= 4!\; \times \;5!$

Note: In such types of questions it is advisable to remember basic concepts of circular permutation that is the number of circular permutations of $n$ different things taken all at a time is $\left( {n - 1} \right)!$, if clockwise or anticlockwise orders are taken as different.