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

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{In how many ways can the letters of the word MISSISSIPPI be rearranged ?}\hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i so did not look that up ;)

steve816
 one year ago
Best ResponseYou've already chosen the best response.3Really? copy and paste?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0please pay attention to word 'rearranged '.

steve816
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{ 11! }{ 4! * 4! * 2! }\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is arranging and rearranging same thing

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.134650 :) yes!! they r same

steve816
 one year ago
Best ResponseYou've already chosen the best response.3@imqwerty Did you use my equation to get the answer?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i m confused very much

steve816
 one year ago
Best ResponseYou've already chosen the best response.3But your rating is 97 and you're a mathlete... I thought you were smart lol

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1yea @steve816 i used the same equation to get the answer :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess, tons of ways? XD

steve816
 one year ago
Best ResponseYou've already chosen the best response.3I'm messing with ya :p

dan815
 one year ago
Best ResponseYou've already chosen the best response.5M, i,i,i,i,S,S,S,S,P,P 1m,4i,4s,2p 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
 one year ago
Best ResponseYou've already chosen the best response.5there is a nice way to picture it

dan815
 one year ago
Best ResponseYou've already chosen the best response.5like lets say for the 4, is you take those apart and put it to the side

dan815
 one year ago
Best ResponseYou've already chosen the best response.5now 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
 one year ago
Best ResponseYou've already chosen the best response.5something very similiar is happening in the 11! case

dan815
 one year ago
Best ResponseYou've already chosen the best response.5i hope thats clear, just think about it urself till it makes sense

dan815
 one year ago
Best ResponseYou've already chosen the best response.5you should clearly underestand, how the choosing formula comes and picking formula comes about

dan815
 one year ago
Best ResponseYou've already chosen the best response.5then all these will feel natural

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@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/permutationsrepeateditems.php
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