The number of ways in which 6 boys and 4 girl | vedclass

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(C) Total number of ways to seat $10$ people around a round table is $(10-1)! = 9! = 362880$. Let the $2$ special boys be $B_1, B_2$ and the special g

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$332640$

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(C) Total number of ways to seat $10$ people around a round table is $(10-1)! = 9! = 362880$. Let the $2$ special boys be $B_1, B_2$ and the special girl be $G_1$. We want to find the number of arrangements where $B_1, B_2, G_1$ do not sit together. Using the complement method,we subtract the cases where they sit together from the total. Treating $(B_1, B_2, G_1)$ as one unit,we have $8$ units to arrange in a circle: $(8-1)! = 7! = 5040$. Within the unit,$B_1, B_2, G_1$ can be arranged in $3! = 6$ ways. Total cases where they sit together = $5040 	imes 6 = 30240$. Number of ways they do not sit together = $362880 - 30240 = 332640$.

The number of ways in which $6$ boys and $4$ girls can be seated around a round table such that $2$ special boys and a special girl never sit together is

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