How many different words can be formed by jum | vedclass

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(D) The word $MISSISSIPPI$ contains $11$ letters: $M=1, I=4, S=4, P=2$.First,arrange the letters other than $S$. These are $M, I, I, I, I, P, P$ ($7$

Vedclass · Accepted Answer

$7 	imes ^6C_4 	imes ^8C_4$

Vedclass · Answer

(D) The word $MISSISSIPPI$ contains $11$ letters: $M=1, I=4, S=4, P=2$.First,arrange the letters other than $S$. These are $M, I, I, I, I, P, P$ ($7$ letters).The number of ways to arrange these $7$ letters is $\frac{7!}{4!2!} = \frac{5040}{24 	imes 2} = \frac{5040}{48} = 105$.There are $8$ possible gaps created by these $7$ letters (including the ends) where the $4$ $S$'s can be placed so that no two $S$ are adjacent.The number of ways to choose $4$ gaps out of $8$ is $^8C_4 = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.The total number of words is $105 	imes 70 = 7350$.Evaluating the options: $^8C_4 = 70$. Option $B$ is $6 	imes 7 	imes 70 = 42 	imes 70 = 2940$. Option $A$ is $8 	imes 15 = 120$. Upon re-evaluating the expression $\frac{7!}{4!2!} 	imes ^8C_4 = 105 	imes 70 = 7350$. The provided options seem to be in a symbolic form. Given the structure,$7 	imes ^6C_4 	imes ^8C_4$ is $7 	imes 15 	imes 70 = 7350$.

How many different words can be formed by jumbling the letters in the word $MISSISSIPPI$ in which no two $S$ are adjacent?

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