'MISSISSIPPI': arrangements of letters with no consec. S's

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pi_pie_lover

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I've been working on this question for a while, and i'm in a slump.

How many arrangments of Mississippi are there with no consecutive S's?

I started with trying to find the ways that it didn't work by grouping all the S's together, as if it were one letter. I then did the equation 10!/(4![*]2!). I was going to continue doing this system, but i'm not sure if this way will work.

Any tips?
 
pi_pie_lover said:
How many arrangments of Mississippi are there with no consecutive S's?
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I think the following will work:

You have eleven "slots" for letters. The M, Is, and Ps may be arranged however you like, but no two (or more) Ss may be side-by-side.

Considering only seven slots, in how many (distinguishable) ways can you arrange the other seven letters? (Use the formula they gave you for this.)

Given all eleven slots, in how many ways can you choose four so that there is at least one empty slot between any given pair of slots? (It might help to think of this as being four double-wide slots, there being an imaginary twelfth slot, and you're needing to pick four of six double-wide slots. Since the first S might go in the first slot, or else it might not, you might want to look at this from each of the right- and left-hand views. That is, multiply the one "six-choose-four" by two.)

Since the ordering (left to right) of the other letters does not affect how you chose the slots for the Ss, the two are independent. So multiply to get the total number of ways.

Have fun! :D

Eliz.
 
Here is another way to think about this problem.
The other letters will stand between the S’s as separators: _M_I_I_I_P_P_I_
The number of ways those can be arranged is: \(\displaystyle \frac {7!} {(4!)(2!)}\).
Now there are eight blanks into which we can place the S’s: \(\displaystyle \binom {8}{4}\).

Answer: \(\displaystyle \binom {8}{4}\frac {7!} {(4!)(2!)}\).
 
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