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mathmath333

  • one year ago

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

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{In how many ways can the letters of the word MISSISSIPPI be rearranged ?}\hspace{.33em}\\~\\ \end{align}}\)

  2. steve816
    • one year ago
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    Easy peasy

  3. anonymous
    • one year ago
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    i so did not look that up ;)

  4. steve816
    • one year ago
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    Really? copy and paste?

  5. anonymous
    • one year ago
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    XD

  6. ganeshie8
    • one year ago
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    lol

  7. mathmath333
    • one year ago
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    please pay attention to word 'rearranged '.

  8. steve816
    • one year ago
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    \[\frac{ 11! }{ 4! * 4! * 2! }\]

  9. steve816
    • one year ago
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    There you go.

  10. mathmath333
    • one year ago
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    is arranging and rearranging same thing

  11. imqwerty
    • one year ago
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    34650 :) yes!! they r same

  12. mathmath333
    • one year ago
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    r u sure

  13. steve816
    • one year ago
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    @imqwerty Did you use my equation to get the answer?

  14. mathmath333
    • one year ago
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    i m confused very much

  15. steve816
    • one year ago
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    But your rating is 97 and you're a mathlete... I thought you were smart lol

  16. imqwerty
    • one year ago
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    yea @steve816 i used the same equation to get the answer :)

  17. steve816
    • one year ago
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    Wow, Mr.Bond

  18. anonymous
    • one year ago
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    I guess, tons of ways? XD

  19. steve816
    • one year ago
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    Brony go away.

  20. anonymous
    • one year ago
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    excuse me?

  21. steve816
    • one year ago
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    I'm messing with ya :p

  22. anonymous
    • one year ago
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    -.-

  23. dan815
    • one year ago
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    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
    • one year ago
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    in 11! ways it* makes

  25. dan815
    • one year ago
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    there is a nice way to picture it

  26. dan815
    • one year ago
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    like lets say for the 4, is you take those apart and put it to the side

  27. dan815
    • one year ago
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    |dw:1441175211779:dw|

  28. dan815
    • one year ago
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    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
    • one year ago
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    |dw:1441175349107:dw|

  30. dan815
    • one year ago
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    something very similiar is happening in the 11! case

  31. dan815
    • one year ago
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    i hope thats clear, just think about it urself till it makes sense

  32. dan815
    • one year ago
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    you should clearly underestand, how the choosing formula comes and picking formula comes about

  33. dan815
    • one year ago
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    then all these will feel natural

  34. mathmate
    • one year ago
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    @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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