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how many different word can be formed by jumbling the letter of the word

MISSISSIPPI in which no three $"S"$ occur together

No. of arrangement of the words MISSISSIPPI is $ = \frac{11!}{4!\cdot 4!\cdot 2!}$

now arrangement of the words in which all $ "S"$ are together is $ = \frac{8!}{4!\cdot 2!}$

total no. of arrangements of the words in which all four $"S"$ are occur together is $ = \frac{11!}{4!\cdot 4!\cdot 2!}-\frac{8!}{4!\cdot 2!}$

i want be able to go further , could some help me with this, Thanks

DXT
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3 Answers3

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All arrangements : $11!/(4! 4! 2!)=34650$

4s together : $8!/4!2!=840 $

3s together 1s apart : $ 56*7!/(4!2!)=5880$

[When $3s=X$ is at the beginnining or at the end $14*7!/(4!2!)$ cases and if not $42*7!/(4!2!)$ cases]

which gives you $27930$ cases ..

next
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Here is a way using stars and bars

Consider the $4$ identical $S's$ ("stars") to be placed in $8$ boxes with a maximum of $3$ in any box,

$\boxed.M\boxed.P\boxed.P\boxed.I\boxed.I\boxed.I\boxed.I\boxed. $

We must exclude any arrangement that has $3$ or more in any of the $8$ boxes,

so applying stars and bars, there are $\binom{11}7- \binom81\binom87 = 266$ ways.

The other letters, which were acting as "bars", can be permuted in $\frac{7!}{2!4!} = 105$ ways,

thus answer $= 266\cdot105 = 27,930$

true blue anil
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    @Next: You were right, I have modified my answer, using a different system. +1 for your correct answer which was arrived at first. – true blue anil Nov 22 '16 at 16:14
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Hint:

Like you counted situation with 4 "S" together, let "SSS" be represented as "one" letter. Then we have 9 spaces in total and:

1) SSS is at the start or end - then we choose 1 out of 7 spots for the last S so it is not near SSS and arrange remaining letters on remaining 7 spots.

2) SSS is in the middle - then we choose 1 out of 6 spots for last S so it is not near SSS and arrange remaining letters on remaining 7 spots.

Hope this helps :)

MaliMish
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