If 6 boys and 3 girls are to be seated on 9 c | vedclass

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(B) Total number of ways to arrange $6$ boys and $3$ girls in $9$ seats is $9!$.For the condition that the end seats are occupied by girls,we choose $

Vedclass · Accepted Answer

$\frac{5}{84}$

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(B) Total number of ways to arrange $6$ boys and $3$ girls in $9$ seats is $9!$.For the condition that the end seats are occupied by girls,we choose $2$ girls out of $3$ for the ends in $^3P_2 = 3 	imes 2 = 6$ ways.The remaining $7$ seats are to be filled by $6$ boys and $1$ girl.To ensure no two girls are side by side,we first arrange the $6$ boys in $6!$ ways.This creates $7$ gaps. Since the $2$ end seats are already occupied by girls,we have $5$ internal gaps available for the remaining $1$ girl.Number of ways to place the last girl is $5$.Total favorable ways = $6 	imes 6! 	imes 5 = 30 	imes 720 = 21600$.Total possible arrangements = $9! = 362880$.Probability = $\frac{21600}{362880} = \frac{5}{84}$.

If $6$ boys and $3$ girls are to be seated on $9$ chairs for a photograph,then the probability that the end seats are occupied by the girls and no two girls are side by side is

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