If the letters of the word ASSASSIN are arran | vedclass

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(D) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(2), N(1)$.Total arrangements = $\frac{8!}{2!4!2!} = \frac{40320}{2 	imes 24 	imes 2} =

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$	ext{None of these}$

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(D) The word $ASSASSIN$ contains $8$ letters: $A(2), S(4), I(2), N(1)$.Total arrangements = $\frac{8!}{2!4!2!} = \frac{40320}{2 	imes 24 	imes 2} = 420$.To ensure no two $S$s occur together,we arrange the remaining letters $A, A, I, N$ first.Number of ways to arrange $A, A, I, N$ = $\frac{4!}{2!} = 12$.These $4$ letters create $5$ gaps: $\_ L \_ L \_ L \_ L \_$.We need to place $4$ $S$s in these $5$ gaps,which can be done in $^5C_4 = 5$ ways.Favorable outcomes = $12 	imes 5 = 60$.Probability = $\frac{60}{420} = \frac{1}{7}$.Note: The provided option $\frac{1}{14}$ is incorrect based on the standard calculation.

If the letters of the word $ASSASSIN$ are arranged in a row at random,what is the probability that no two $S$s occur together?

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