The number of ways, in which 5 girls and 7 boys can be seated at a...

Given,

$7$ boys and $5$ girls are to be seated around a circular such that no two girls to be seated together,

Now we know that $n$ objects can be arranged in a circle in $( n - 1 ) !$ ways.

Let us first arrange $7$ boys in circular arrangement in \left(7 - 1\right) !$ways.</p><p>Now there will be$7$gaps.</p><p>So let us select any$5$gaps out of$7$gaps and arrange$5$girls in the chosen gaps. This can be done in$C5 7 \times 5 !$ways.</p><p>Hence, required arrangements are$6 ! \times C5 7 \times 5 !

$= 6 \times 5 ! \times$\dfrac{7 \times 6}{2}$\times 5 !</p><p>$= 126{\left(5 !\right)}^{2}$.</p><p>Therefore, required arrangements are$126{\left(5 !\right)}^{2}