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Math Help - Binary combination/permutation help

  1. #1
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    Binary combination/permutation help

    Hello helpful smart community

    i have some questions involving permutation/combinations and need your help.

    here it goes:

    im tring to find out how many different arrangements of 4 (1's) can be put in 40 (0's)

    ex:
    : 0000000000000000000000000000000000000000 empty
    1 : 1111000000000000000000000000000000000000
    2 : 0000000000000000000000000000000000001111
    3 : 0000010001000000000000001000000000000100
    etc....

    how do i calculate that? O.o

    also what is the chance of finding specific one of those in 1000000 random attempts.

    any idea's?
    Last edited by zipzaaaap6656; November 7th 2011 at 09:37 PM.
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  2. #2
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    Re: Binary combination/permutation help

    Your task can be rephrased to distributing 35 objects into 5 buckets. {5+35-1} \choose {5-1}
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  3. #3
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    Re: Binary combination/permutation help

    hmmmm thanks for reply but im getting the feeling thats not it ^^
    can you explain more please?

    i think i found out how to do it, but can someone explain to me how to calculate it exactly?

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    Last edited by zipzaaaap6656; November 7th 2011 at 09:55 PM.
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  4. #4
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    Re: Binary combination/permutation help

    Yours is the "Empty Urns are Allowed" case.
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  5. #5
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    Re: Binary combination/permutation help

    Hellom, zipzaaaap6656!

    (a) How many different arrangements are there for four 1's and forty 0's?

    We have a 44-digit number.
    The four 1's can be placed in . {44\choose4} \,=\,135,\!751 ways.
    The forty 0's are placed in the remaining positions.

    Therefore, there are 135,751 possible arrangements.




    (b) What is the probability of finding a specific number in 1,000,000 random attempts.?

    I'll assume that the selection is made with replacement.

    P(\text{not-number}) \:=\:\frac{135,\!750}{135,\!751}

    P(\text{not-number in 1,000,000 tries}) \:=\:\left(\frac{135,\!750}{135,\!751}\right)^{1,0  00,000} \;=\;0.000\: 632\, 104

    P(\text{number at least once in 1,000,000 tries})
    . . . . . . \;=\;1 - 0.000\:632\,104 \;=\;0.999\,367\,896

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