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Math Help - how to determin number of combinations

  1. #1
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    how to determin number of combinations

    hello...

    I'm having a bit of problem with one simple task First sorry if I put this thread in wrong place

    I know how to calculate number of combinations (without repetition)

    \frac {n!}{k!(n-k)!}

    and that's ok

    but now I have big problem I have to calculate number of combinations with repetitions so i tried like this

    when i have 2 variables A and B with second expansion i have :

    AA, AB, BA, BB

    so it's four combinations

    for third expansion i have :

    AAA, AAB, ABA, BAA, BBB, BBA, BAB, ABB

    so there is 8 combinations .... and at first i think formula should be n^k or in my case now 2^n meaning 2^2 = 4 and 2^3 = 8 but as I get to problem with 4 variables and need to do third expansion by that formula I need 64 combinations... I wrote all of them and I got 40... now I'm confused very much because in next task I have problem with 10 variables and third expansion ... and there is no way that i can wrote them all to find pattern ....

    can anyone please help me with this I also find some formulas for similar things but none of them can be any use to me in this problem
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  2. #2
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    here is how i got combinations with 4 variables and third expansion

    \begin{matrix}<br />
AAA & BBB & CCC & DDD & AAC & BBD\\ <br />
AAB & BBC & CCD & DDA & ACA & BDB\\ <br />
ABA & BCB & CDC & DAD & CAA & DBB\\ <br />
BAA & CBB & DCC & ADD & ACC & BDD\\ <br />
ABB & BCC & CDD & DAA & CAC & DBD\\ <br />
BAB & CBC & DCD & ADA & CCA & DDB\\ <br />
BBA & CCB & DDC & AAD &  & <br />
\end{matrix}

    so there are 40 combinations
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  3. #3
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    To be honest I don’t really follow your question.
    I think you mean if there are ten symbols, ABCDEFGHIJ, how many strings of length ten can be formed using any or all of the ten.

    For example: FJBBIIAGHI\text{ or }JIHGFEDCBA are two possible strings.
    In that example there are 10^{10} possible strings.

    In general, if you have \mathif{N} different symbols the number of possible strings of length \mathif{K}, using any or all of the symbols, is \mathif{N}^{\mathif{K}}

    If you mean something else, please try to make it clearer.
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  4. #4
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    sorry for bad explanation...
    It's more like how much combinations can U wrote with 10 symbols in string with length of 3 with repetitions
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  5. #5
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    Quote Originally Posted by sedam7 View Post
    sorry for bad explanation...
    It's more like how much combinations can U wrote with 10 symbols in string with length of 3 with repetitions
    The answer to that question is just 10^{3}=10\cdot 10\cdot 10.

    In the case of four symbols it is 4^3=64.
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  6. #6
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    Quote Originally Posted by sedam7 View Post
    hello...

    I'm having a bit of problem with one simple task First sorry if I put this thread in wrong place

    I know how to calculate number of combinations (without repetition)

    \frac {n!}{k!(n-k)!}

    and that's ok

    but now I have big problem I have to calculate number of combinations with repetitions so i tried like this

    when i have 2 variables A and B with second expansion i have :

    AA, AB, BA, BB

    so it's four combinations

    for third expansion i have :

    AAA, AAB, ABA, BAA, BBB, BBA, BAB, ABB

    so there is 8 combinations .... and at first i think formula should be n^k or in my case now 2^n meaning 2^2 = 4 and 2^3 = 8 but as I get to problem with 4 variables and need to do third expansion by that formula I need 64 combinations... I wrote all of them and I got 40... now I'm confused very much because in next task I have problem with 10 variables and third expansion ... and there is no way that i can wrote them all to find pattern ....

    can anyone please help me with this I also find some formulas for similar things but none of them can be any use to me in this problem
    If you have 2 things (as letters) and n places to put them (as in 2 or three letter words) then you have 2 choices n times. The number of possible "words" is 2^n

    More generally, if you have m objects and n places to put them, there are m^n possible combinations.
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  7. #7
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    Quote Originally Posted by Plato View Post
    The answer to that question is just 10^{3}=10\cdot 10\cdot 10.

    In the case of four symbols it is 4^3=64.
    hehehehehe... yes  4^3 = 64 but up there in post #2 i wrote all of them and there are just 40 combinations or i made mistake
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  8. #8
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    Quote Originally Posted by sedam7 View Post
    hehehehehe... yes  4^3 = 64 but up there in post #2 i wrote all of them and there are just 40 combinations or i made mistake
    You made a mistake.
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  9. #9
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    Quote Originally Posted by Plato View Post
    You made a mistake.
    LOL... I'm idiot i just do it with 2 different variables ....


    sorry for all mess that i make .... and thank you all very very much
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