In how many ways can the letters of the word | vedclass

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(D) The word $ASSASSIN$ contains $9$ letters: $A(2), S(4), I(2), N(1)$.First,arrange the letters other than $S$,which are $A, A, I, I, N$. The number

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(D) The word $ASSASSIN$ contains $9$ letters: $A(2), S(4), I(2), N(1)$.First,arrange the letters other than $S$,which are $A, A, I, I, N$. The number of ways to arrange these $5$ letters is $\frac{5!}{2!2!} = \frac{120}{4} = 30$.These $5$ letters create $6$ gaps (including ends): $\_ A \_ A \_ I \_ I \_ N \_$.We need to place $4$ $S$s in these $6$ gaps so that no two $S$s are together. The number of ways to choose $4$ gaps out of $6$ is $^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$.Total number of arrangements = $30 	imes 15 = 450$.

In how many ways can the letters of the word $ASSASSIN$ be arranged so that no two $S$s occur together?

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