## mathmath333 one year ago In how many ways can the letters of the word MISSISSIPPI be rearranged ?

1. 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}}

2. steve816

Easy peasy

3. anonymous

i so did not look that up ;)

4. steve816

Really? copy and paste?

5. anonymous

XD

6. ganeshie8

lol

7. mathmath333

please pay attention to word 'rearranged '.

8. steve816

$\frac{ 11! }{ 4! * 4! * 2! }$

9. steve816

There you go.

10. mathmath333

is arranging and rearranging same thing

11. imqwerty

34650 :) yes!! they r same

12. mathmath333

r u sure

13. steve816

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

14. mathmath333

i m confused very much

15. steve816

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

16. imqwerty

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

17. steve816

Wow, Mr.Bond

18. anonymous

I guess, tons of ways? XD

19. steve816

Brony go away.

20. anonymous

excuse me?

21. steve816

I'm messing with ya :p

22. anonymous

-.-

23. 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!)

24. dan815

in 11! ways it* makes

25. dan815

there is a nice way to picture it

26. dan815

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

27. dan815

|dw:1441175211779:dw|

28. 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

29. dan815

|dw:1441175349107:dw|

30. dan815

something very similiar is happening in the 11! case

31. dan815

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

32. dan815

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

33. dan815

then all these will feel natural

34. 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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