How many words can be formed by arranging the | vedclass

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Question

(A) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.First,we arrange the letters other than $S$,which are $M, I, I, I, I, P, P$

Vedclass · Accepted Answer

$7 	imes ^6C_4 	imes ^8C_4$

Vedclass · Answer

(A) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.First,we arrange the letters other than $S$,which are $M, I, I, I, I, P, P$. There are $7$ such letters.The number of ways to arrange these $7$ letters is $\frac{7!}{4! 	imes 2!} = \frac{7 	imes 6 	imes 5}{2} = 105$.Note that $105 = 7 	imes 3 	imes 5 = 7 	imes ^6C_4$.Now,to ensure no two $S$ are together,we place the $4$ $S$ letters in the gaps created by the $7$ letters. There are $8$ such gaps (including ends).The number of ways to choose $4$ gaps out of $8$ is $^8C_4$.Thus,the total number of arrangements is $7 	imes ^6C_4 	imes ^8C_4$.

How many words can be formed by arranging the letters of the word $MISSISSIPPI$ such that no two $S$ are together?

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