The number of ways in which 5 girls and 3 boy | vedclass

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(D) To ensure no two boys are together,we first arrange the $5$ girls in a row. The number of ways to arrange $5$ girls is $5! = 120$.There are $6$ po

Vedclass · Accepted Answer

$14400$

Vedclass · Answer

(D) To ensure no two boys are together,we first arrange the $5$ girls in a row. The number of ways to arrange $5$ girls is $5! = 120$.There are $6$ possible gaps (including the ends) created by the $5$ girls: $\_ G_1 \_ G_2 \_ G_3 \_ G_4 \_ G_5 \_$.We need to place $3$ boys in these $6$ gaps. The number of ways to choose $3$ gaps out of $6$ is $^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.The $3$ boys can be arranged in these $3$ chosen gaps in $3! = 6$ ways.Therefore,the total number of ways is $5! 	imes ^{6}C_{3} 	imes 3! = 120 	imes 20 	imes 6 = 14400$.

The number of ways in which $5$ girls and $3$ boys can be seated in a row so that no two boys are together is:

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