In how many ways can $5$ girls and $3$ boys be seated in a row so that no two boys are together?
In how many ways can $5$ girls and $3$ boys be seated in a row so that no two boys are together?
Let us first seat the $5$ girls. This can be done in $5 !$ ways. For each such arrangement, the three boys can be seated only at the cross marked places.
$\times G \times G \times G \times G \times G \times$
There are $6$ cross marked places and the three boys can be seated in $^{6} P _{3}$ ways. Hence, by multiplication principle, the total number of ways
${ = 5!{ \times ^6}{P_3} = 5! \times \frac{{6!}}{{3!}}}$
${ = 4 \times 5 \times 2 \times 3 \times 4 \times 5 \times 6 = 14400}$
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