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(A) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.The total number of distinct permutations is given by $\frac{11!}{4!4!2!} =

Vedclass · Accepted Answer

$33810$

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(A) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.The total number of distinct permutations is given by $\frac{11!}{4!4!2!} = \frac{39916800}{24 	imes 24 	imes 2} = \frac{39916800}{1152} = 34650$.To find the number of permutations where the four $I$'s come together,we treat the block $(IIII)$ as a single object. Now we have $8$ objects: $(IIII), M, S, S, S, S, P, P$.The number of ways to arrange these $8$ objects,where $S$ repeats $4$ times and $P$ repeats $2$ times,is $\frac{8!}{4!2!} = \frac{40320}{24 	imes 2} = \frac{40320}{48} = 840$.The number of permutations where the four $I$'s do not come together is the total permutations minus the permutations where they are together:$34650 - 840 = 33810$.

In how many of the distinct permutations of the letters in $MISSISSIPPI$ do the four $I$'s not come together?

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