In how many ways can 5 boys and 5 girls stand | vedclass

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(C) To ensure no two girls stand together,we first arrange the $5$ boys in a row. The number of ways to arrange $5$ boys is $5!$.After arranging the b

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$5! 	imes 6!$

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(C) To ensure no two girls stand together,we first arrange the $5$ boys in a row. The number of ways to arrange $5$ boys is $5!$.After arranging the boys,there are $6$ possible gaps (including the ends) where the $5$ girls can be placed: _ $B_1$ _ $B_2$ _ $B_3$ _ $B_4$ _ $B_5$ _.The number of ways to choose $5$ gaps out of $6$ and arrange the $5$ girls in them is given by $^6P_5$.Since $^6P_5 = \frac{6!}{(6-5)!} = 6!$,the total number of arrangements is $5! 	imes 6!$.

In how many ways can $5$ boys and $5$ girls stand in a row so that no two girls may be together?

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