3 boys B_1, B_2, B_3 and 6 girls G_1, G_2, . | vedclass

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(B) Total number of people $= 3 + 6 = 9$.Let $S$ be the set of all arrangements where $B_1, B_2$ are separated.Let $A$ be the condition that $B_1, B_2

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$44 	imes 7!$

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(B) Total number of people $= 3 + 6 = 9$.Let $S$ be the set of all arrangements where $B_1, B_2$ are separated.Let $A$ be the condition that $B_1, B_2$ are together.Let $B$ be the condition that $G_1, G_2$ are together.We want to find the number of arrangements where $B_1, B_2$ are separated $AND$ $G_1, G_2$ are separated.This is given by: (Total arrangements where $B_1, B_2$ are separated) - (Arrangements where $B_1, B_2$ are separated $AND$ $G_1, G_2$ are together).$1$. Total arrangements where $B_1, B_2$ are separated: Total arrangements - Arrangements where $B_1, B_2$ are together $= 9! - (8! 	imes 2!) = 9 	imes 8! - 2 	imes 8! = 7 	imes 8!$.$2$. Arrangements where $B_1, B_2$ are separated $AND$ $G_1, G_2$ are together: (Arrangements where $G_1, G_2$ are together) - (Arrangements where $G_1, G_2$ are together $AND$ $B_1, B_2$ are together).Arrangements where $G_1, G_2$ are together $= 8! 	imes 2!$.Arrangements where $G_1, G_2$ are together $AND$ $B_1, B_2$ are together $= 7! 	imes 2! 	imes 2!$.So,(Arrangements where $B_1, B_2$ are separated $AND$ $G_1, G_2$ are together) $= (8! 	imes 2!) - (7! 	imes 4) = (16 	imes 7!) - (4 	imes 7!) = 12 	imes 7!$.Required ways $= (7 	imes 8!) - (12 	imes 7!) = (56 	imes 7!) - (12 	imes 7!) = 44 	imes 7!$.

$3$ boys $B_1, B_2, B_3$ and $6$ girls $G_1, G_2, . . . , G_6$ are to be seated in a row. The number of ways they can be seated such that $B_1, B_2$ are separated and $G_1, G_2$ are also separated is equal to:

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