The number of ways in which 5 boys and 3 girl | vedclass

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(B) Total number of people $= 5 + 3 = 8$. Total ways to arrange $8$ people on a round table $= (8-1)! = 7!$. Now,consider the case where $B_1$ and $G_

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$5 	imes 6!$

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(B) Total number of people $= 5 + 3 = 8$. Total ways to arrange $8$ people on a round table $= (8-1)! = 7!$. Now,consider the case where $B_1$ and $G_1$ sit together. Treat $(B_1G_1)$ as one unit. Now we have $7$ units to arrange on a round table,which can be done in $(7-1)! = 6!$ ways. Within the unit,$B_1$ and $G_1$ can be arranged in $2! = 2$ ways. So,the number of ways they sit together $= 6! 	imes 2$. The number of ways they never sit adjacent $= 7! - (2 	imes 6!) = 7 	imes 6! - 2 	imes 6! = (7-2) 	imes 6! = 5 	imes 6!$.

The number of ways in which $5$ boys and $3$ girls can be seated on a round table,if a particular boy $B_1$ and a particular girl $G_1$ never sit adjacent to each other,is

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