How many different words can be found by jumbling the letters in the word' MISSISSIPPI'in which No 2 S are adjacent?

Mississippi has 11 letters, 4 of which are S. We don't want the Ss to be adjacent to each other. So the first thing to figure out is the number of ways to arrange the Ss so that they aren't adjacent.

#1,3,5,7#
#1,3,5,8#
#1,3,5,9#
#1,3,5,10#
#1,3,5,11#

#1,3,6,8#
#1,3,6,9#
#1,3,6,10#
#1,3,6,11#

#1,3,7,9#
#1,3,7,10#
#1,3,7,11#

#1,3,8,10#
#1,3,8,11#

#1,3,9,11#

This is 15 so far. Notice that for "each step over" of the third digit, we lost one possibility (ending up with #5+4+3+2+1=15#)

#1,4,6,8#
#1,4,6,9#
#1,4,6,10#
#1,4,6,11#

And let's stop here. Notice that we've moved the "second digit" over one and we have 4 possibilities. Then when we move the "third digit", we'll lose 1 possibility each time, which will give #4+3+2+1=10#

This means that #1,5,...# gives 6, #1,6,...# gives 3, #1,7,...# gives 1

Or all the possibilities for the staring digit 1 is #15+10+6+3+1=35#

For starting digit 2, we'll get:

#2,4,6,8#
#2,4,6,9#
#2,4,6,10#
#2,4,6,11#

So #2,4,...# gives 4 meaning we'll get 20 #(10+6+3+1)# total from the 2s.

We'll get 10 ways from starting digit 3.
We'll get 4 ways from starting digit 4.
And we'll get 1 way from starting digit 5.

This all gives:

#35+20+10+4+1=70# ways to arrange the S's.

(There's probably a way to find this much quicker and faster using partitions but I'm not seeing that solution at the moment).

In and amongst the S's will be the other letters. 7 of them in fact, but we have duplicates: 4 of the I and 2 of the P. If they were all unique, we'd say there #7!# ways to arrange them (for each arrangement of the S's), but we need to divide out by the way each group can arrange internally, which is the number of that letter, factorial.

All this gives:

#70xx(7!)/(4!2!)=70xx(7xx3xx5)=7350# different arrangements of letters