Question

This is a stats question, but I'm hoping someone will be able to help.

how many different ways to arrange the letters in Mississippi are there assuming letters are not distinct and no two I's can be next to each other?
The answer is 8 choose 4x (7!/4!2!1!).
I understand 7!/4!2!1! is the ways to rearrange the letters that are not I, and that 8 choose 4 is the ways to arrange I so that no two are next to each other. I don't understand how they come up with 8 choose 4.

Answer 1

In general:

The number of ways to arrange letters:

\[\frac{\text{no. of letters}!}{\text{no of each letter repeats}!}\]

Answer 2

To Find No of Arrangment in whhich No two i's come Togeter is

Total - no of arrg in which two i's come together

Answer 3

Did u Understand @michellew20

Answer 4

Hero, you are correct, but then I need to find the # of ways the I's can be arranged so that they will not be next to each other.
Yahoo, I would do it that way, but I don't know how.

Answer 5

Take two i's as one.....

Answer 6

\[\frac{ 10! }{ 2!*4!*2! } * \frac{ 2! }{ 2! }\]

Answer 7

oh, wouldn't it be 11!/(4!4!2!1!)- 9!/(2!4!2!1!)?

Answer 8

cause 9 is the # of letters if you think of ii as 1 letter, but this still does not cover the situation where ii and ii are next to each other.

Answer 9

11!/(4!4!2!1!)- 10!/(2!4!2!1!)?

Answer 10

taking 2 i's as one will reduce 1 the total so 11- 1 = 10

Answer 11

but there are 4 I's, so wouldn't there be 2 less?

Answer 12

@michellew20

Here's a similar problem involving Mississippi and the letters p that do not stand together. See example 2.

http://www.mymathcounts.com/documents/Chapter24Combinatotics1.pdf