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