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Math Help - How many possible arrangements ...

  1. #1
    Super Member Bacterius's Avatar
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    How many possible arrangements ...

    Hello,
    I have a question

    I have been trying to investigate the formula that gives the number of possible arrangements of n elements in m cells. But I just can't get this formula. I've been thinking a lot and I think it involves factorials but this is just confusing me since two different arrangements could be equivalent given other formulas

    For example, if I have three cells ( m = 3), I can arrange two elements ( n = 2) in three different combinations.

    Can anyone help me out on this formula (it's obviously very simple as I cannot find it ...) ? Thanks all

    PS : I have to apply the formula to n = 8 and m = 256 next. Seems like I'm going to have to count decimal digits lol.
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  2. #2
    MHF Contributor
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    Hi Bacterius,

    Are you working in binary or generally?

    Mp_N=\frac{2^8!}{(2^8-2^3)!} for distinguishable cells.

    Are you looking for something beyond that?
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  3. #3
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Hi Bacterius,

    Are you working in binary or generally?

    Mp_N=\frac{2^8!}{(2^8-2^3)!} for distinguishable cells.

    Are you looking for something beyond that?
    Hi Archie Meade,
    well I was trying to find the formula you just gave me (thanks! ) for a table I had to build, but I thought why not finding the general formula (which I can now deduce from your answer).

    Thanks again
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  4. #4
    MHF Contributor
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    Hi Bacterius,

    the "permutations" or arrangements formula will do that

    Arrange all N cells.... N!
    Arrange N cells from a total of M cells Mp_N=\frac{M!}{(M-N)!}

    This formula is just a "tweaking" of the actual situation.

    For example, arrange 9 distinguishable cells in all orders...

    9(8)7)6(5)4(3)2=9!

    arrange only 4 of the 9...

    9(8)7(6) just happens to be the same as 9! with the 5! missing
    and the 5! can be dispensed with by dividing 9! by 5!
    5 is the difference between 9 and 4

    so, 9(8)7(6)=\frac{9!}{5!}=\frac{9!}{(9-4)!}

    hence the situation of arranging n from m can be formulated from the M and N using

    Mp_N=\frac{M!}{(M-N)!}
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