Arrangement of word MISSISSIPPI in which no three S occur together 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
 A: 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 ..
A: 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 $= 266\cdot105 = 27,930$
A: 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 :)
