Question

In how many distinct permutations of the letter in "MISSISSIPPI" do the four I's not come together?​

Answer 1

First find the

total no of permutation possible=(11!/4!4!2!)

=34650

As 's' and 'i' are repeated 4 times and 'p' is repeated 2 times, that is why 11! is divided by 4!x4!x2!

Now assume the 4 i's together as one word

The no. of letters left is 8.

1-m 4-s 2-p and 1-containing 4 i's.

The permutation for 4 i's together will be= 8!/(4!2!)

=840

We want no. of permutation in which 4i's are not together

P= Total permutation - permutation of 4 i's together

After solving

P=33810

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Answer 2

${ \underline{ \huge{ \mathtt{ \: \fbox{ Solution : }\: \: }}}}$

There are total 11 letters , of which I occurs 4 times , S occurs 4 times and P occurs 2 times

Therefore ,

The total permutations (without any restriction) is

$\sf = \frac{11!}{4! \times 2! \times 4!} \\ \\ \sf = \frac{11 \times 10 \times 9 \times \cancel 8 \times 7 \times \cancel6 \times 5 \times \cancel{4!}}{ \cancel{4 !}\times 2 \times 1 \times \cancel4 \times \cancel3 \times 2 \times 1} \\ \\ \sf = \frac{11 \times 10 \times 9 \times \cancel2 \times 7 \times \cancel2 \times 5}{ \cancel2 \times \cancel{2}} \\ \\ \sf = 34650 \: ways$

Hence , the number of permutations (without any restriction) is 34650 ways

Assume the 4 I's together as a one letter , So now , the number of total letters is 8

Therefore , The number of permutations when all 4 I's occur together is

$\sf = \frac{8}{4 \times 2} \\ \\ \sf = \frac{\cancel8 \times 7 \times 6 \times 5 \times \cancel4}{\cancel4 \times \cancel2 \times 1} \\ \\ \sf = 4 \times 7 \times 6 \times 5 \\ \\ \sf = 840 \: ways$

Hence , the number of permutations when all I's occur together is 840 ways

Now ,

The required number of arrangements = the total permutations (without any restriction) - the number of permutations where all I occur together

Substitute the values , we get

$\sf \hookrightarrow The \: required \: number \: of \: arrangements =34650 - 840 \\ \\ \sf \hookrightarrow The \: required \: number \: of \: arrangements =33810 \: ways$

Hence , the number of permutations when all I's not occur together is 33810 ways