In how many ways can the letters of the word MISSISSIPPI be rearranged ?

- mathmath333

In how many ways can the letters of the word MISSISSIPPI be rearranged ?

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- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{In how many ways can the letters of the word MISSISSIPPI be rearranged ?}\hspace{.33em}\\~\\
\end{align}}\)

- steve816

Easy peasy

- anonymous

i so did not look that up ;)

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## More answers

- steve816

Really? copy and paste?

- anonymous

XD

- ganeshie8

lol

- mathmath333

please pay attention to word 'rearranged '.

- steve816

\[\frac{ 11! }{ 4! * 4! * 2! }\]

- steve816

There you go.

- mathmath333

is arranging and rearranging same thing

- imqwerty

34650 :) yes!! they r same

- mathmath333

r u sure

- steve816

@imqwerty Did you use my equation to get the answer?

- mathmath333

i m confused very much

- steve816

But your rating is 97 and you're a mathlete...
I thought you were smart lol

- imqwerty

yea @steve816 i used the same equation to get the answer :)

- steve816

Wow, Mr.Bond

- anonymous

I guess, tons of ways? XD

- steve816

Brony go away.

- anonymous

excuse me?

- steve816

I'm messing with ya :p

- anonymous

-.-

- dan815

M, i,i,i,i,S,S,S,S,P,P
1-m,4-i,4-s,2-p
total = 1+4+4+2=11
in 11! ways if makes a distinction between the repeated letter cases so we have a multiple of those according to how many redundant cases we got
so 11!/(4!*4!*2!)

- dan815

in 11! ways it* makes

- dan815

there is a nice way to picture it

- dan815

like lets say for the 4, is you take those apart and put it to the side

- dan815

|dw:1441175211779:dw|

- dan815

now if those is are seen as different , and we have to play it in the front like that then thers another 4! ways for every one of those 7! arragenemtents on the right

- dan815

|dw:1441175349107:dw|

- dan815

something very similiar is happening in the 11! case

- dan815

i hope thats clear, just think about it urself till it makes sense

- dan815

you should clearly underestand, how the choosing formula comes and picking formula comes about

- dan815

then all these will feel natural

- mathmate

@mathmath333
This is a problem of "permutation with repeats".
The basic permutation problem is n! arrangements for n distinct objects.
If we have three identical objects, to adjust for the overcount, we have to divide by the number of permutations of the three objects, or 3! .
So in this case, with 4i, 4S, 2p and 1m (total 11), the number of permutations is
\(\Large \frac{11!}{4!4!2!1!}\)
If we remember that the sum of the numbers in the denominator must equal the numerator (4+4+2+1=11), then we won't miss anything.
For example, the same principle applied to 4 distinct objects would give 4!/(1!1!1!1!)=4! that we are familiar with.
You can also read:
http://www.mathwarehouse.com/probability/permutations-repeated-items.php

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