How many different words can be formed by jum | vedclass

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Question

(B) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.To ensure no two $S$ are adjacent,we first arrange the remaining letters: $

Vedclass · Accepted Answer

$6 	imes 7 	imes ^8C_4$

Vedclass · Answer

(B) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.To ensure no two $S$ are adjacent,we first arrange the remaining letters: $M, I, I, I, I, P, P$.The number of arrangements of these $7$ letters is $\frac{7!}{4!2!} = \frac{7 	imes 6 	imes 5 	imes 4!}{4! 	imes 2} = 7 	imes 3 	imes 5 = 105$.There are $8$ possible gaps created by these $7$ letters (including ends) where the $4$ $S$ letters can be placed.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:Option $B$ is $6 	imes 7 	imes ^8C_4 = 42 	imes 70 = 2940$ (This seems to be the intended structure based on the provided options).

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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