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Math Help - Counting question?

  1. #1
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    Counting question?

    Greetings,

    I have the following question in couting. It might be straight forward but I could not find a formula for the answer.

    Let N and m be positive integers.

    How many combinations are there of the form:

    n1, n2, ..., nN such that

    n1 >=0, n2>=0, ..., nN >= 0, and
    n1 + n2 + n3 + ... + nN = m.

    Example: N = 3, m = 4:
    All combinations are given by
    1) 4, 0, 0,
    2) 3, 1, 0,
    3) 3, 0, 1,
    4) 2, 2, 0,
    5) 2, 1, 1,
    6) 2, 0, 1,
    7) 1, 3, 0,
    8) 1, 2, 1,
    9) 1, 1, 2,
    10) 1, 0, 3,
    11) 0, 4, 0,
    12) 0, 3, 1,
    13) 0, 2, 2,
    14) 0, 1, 3,
    15) 0, 0, 4

    Therefore, the number of combinations for this example is 15.

    Best regards.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by sakhr98 View Post
    Greetings,

    I have the following question in couting. It might be straight forward but I could not find a formula for the answer.

    Let N and m be positive integers.

    How many combinations are there of the form:

    n1, n2, ..., nN such that

    n1 >=0, n2>=0, ..., nN >= 0, and
    n1 + n2 + n3 + ... + nN = m.

    Example: N = 3, m = 4:
    All combinations are given by
    1) 4, 0, 0,
    2) 3, 1, 0,
    3) 3, 0, 1,
    4) 2, 2, 0,
    5) 2, 1, 1,
    6) 2, 0, 1,
    7) 1, 3, 0,
    8) 1, 2, 1,
    9) 1, 1, 2,
    10) 1, 0, 3,
    11) 0, 4, 0,
    12) 0, 3, 1,
    13) 0, 2, 2,
    14) 0, 1, 3,
    15) 0, 0, 4

    Therefore, the number of combinations for this example is 15.

    Best regards.
    If I've understood correctly your question, the formula is :

    {m \choose N+m-1}={N-1 \choose N+m-1}

    I can try to explain..
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  3. #3
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    Thank you for the prompt reply.

    I am not sure I get your answer. For the example I put, N = 3 and m = 4, substituting in your formula results in C(2, 6) = C(4, 6).
    But what is the definition of C(2,6)?

    Best regards, Ashraf.
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  4. #4
    Moo
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    The definition ?

    C(p,n)={p \choose n}=\frac{n!}{(n-p)!p!}

    Is it what you want ?
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  5. #5
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    I apologize for my hastiness. Your answer is correct except for the typo where the answer is C(N+m-1, m) = C(N+m-1, N-1).

    Thank you very much.
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  6. #6
    Moo
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    Quote Originally Posted by sakhr98 View Post
    I apologize for my hastiness. Your answer is correct except for the typo where the answer is C(N+m-1, m) = C(N+m-1, N-1).

    Thank you very much.
    Yeah, I have real difficulties with this formula...

    You're welcome
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hello,



    If I've understood correctly your question, the formula is :

    {m \choose N+m-1}={N-1 \choose N+m-1}

    I can try to explain..
    Ahem.

    Moo, I believe the usual convention on the notation for combinations is that \binom{n}{m} is the number of combinations of n objects taken m at a time.

    Therefore, your formula should read
    \binom{N+m-1}{m} = \binom{N+m-1}{N-1}.

    jw
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