If the letters of the word "ASSASSINATION" ar | vedclass

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(C) The word "$ASSASSINATION$" contains $13$ letters: $A(3), S(4), I(2), N(2), T(1)$.Total number of arrangements $= \frac{13!}{3!4!2!2!}$.To ensure n

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$\frac{15}{26}$

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(C) The word "$ASSASSINATION$" contains $13$ letters: $A(3), S(4), I(2), N(2), T(1)$.Total number of arrangements $= \frac{13!}{3!4!2!2!}$.To ensure no two $A$'s come together,we arrange the remaining $10$ letters $(S, S, S, S, I, I, N, N, T)$ first.The number of ways to arrange these $10$ letters is $\frac{10!}{4!2!2!}$.These $10$ letters create $11$ gaps (including ends) where the $3$ $A$'s can be placed.The number of ways to choose $3$ gaps out of $11$ is $^{11}C_3 = \frac{11 	imes 10 	imes 9}{3 	imes 2 	imes 1} = 165$.Required probability $= \frac{\frac{10!}{4!2!2!} 	imes ^{11}C_3}{\frac{13!}{3!4!2!2!}} = \frac{10! 	imes 165 	imes 3!}{13!} = \frac{165 	imes 6}{13 	imes 12 	imes 11} = \frac{165 	imes 6}{1716} = \frac{990}{1716} = \frac{15}{26}$.

If the letters of the word "$ASSASSINATION$" are arranged at random in a row,then the probability that no two $A$'s come together is equal to

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