mathmath333
  • mathmath333
In how many ways can the letters of the word MISSISSIPPI be rearranged ?
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

mathmath333
  • mathmath333
In how many ways can the letters of the word MISSISSIPPI be rearranged ?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathmath333
  • 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
  • steve816
Easy peasy
anonymous
  • anonymous
i so did not look that up ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

steve816
  • steve816
Really? copy and paste?
anonymous
  • anonymous
XD
ganeshie8
  • ganeshie8
lol
mathmath333
  • mathmath333
please pay attention to word 'rearranged '.
steve816
  • steve816
\[\frac{ 11! }{ 4! * 4! * 2! }\]
steve816
  • steve816
There you go.
mathmath333
  • mathmath333
is arranging and rearranging same thing
imqwerty
  • imqwerty
34650 :) yes!! they r same
mathmath333
  • mathmath333
r u sure
steve816
  • steve816
@imqwerty Did you use my equation to get the answer?
mathmath333
  • mathmath333
i m confused very much
steve816
  • steve816
But your rating is 97 and you're a mathlete... I thought you were smart lol
imqwerty
  • imqwerty
yea @steve816 i used the same equation to get the answer :)
steve816
  • steve816
Wow, Mr.Bond
anonymous
  • anonymous
I guess, tons of ways? XD
steve816
  • steve816
Brony go away.
anonymous
  • anonymous
excuse me?
steve816
  • steve816
I'm messing with ya :p
anonymous
  • anonymous
-.-
dan815
  • 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
  • dan815
in 11! ways it* makes
dan815
  • dan815
there is a nice way to picture it
dan815
  • dan815
like lets say for the 4, is you take those apart and put it to the side
dan815
  • dan815
|dw:1441175211779:dw|
dan815
  • 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
  • dan815
|dw:1441175349107:dw|
dan815
  • dan815
something very similiar is happening in the 11! case
dan815
  • dan815
i hope thats clear, just think about it urself till it makes sense
dan815
  • dan815
you should clearly underestand, how the choosing formula comes and picking formula comes about
dan815
  • dan815
then all these will feel natural
mathmate
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.