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Thread: Number of solutions to an equation.

  1. #1
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    Number of solutions to an equation.

    Question:
    Argue that there are exactly $\displaystyle \binom{r}{k}\binom{n-1}{n-r+k}$ solutions of $\displaystyle x_{1}+x_{2}+\cdots+x_{r}=n$ for which exactly $\displaystyle k$ of the $\displaystyle x_{i}$ are equal to $\displaystyle 0$.

    My Logic:
    I started by interpreting $\displaystyle \binom{r}{k}$ to be the number of ways you can select $\displaystyle k$ out of the $\displaystyle r$ terms to be zero. But, I can't figure out what \binom{n-1}{n-r+k} does. I think the $\displaystyle n-1$ part is essentially saying you line up $\displaystyle n$ ones and create a spot between each one where you can insert a bar. I.e. if $\displaystyle n=8$ you have something like:

    1 | 1 | 1 | 1 | 1 | 1 | 1 | 1

    Where the |'s are possible places to insert a bar. In which case it seems to me that you would want to choose $\displaystyle r-k-1$ bars giving you $\displaystyle \binom{n-1}{r-k-1}$. Can someone please point out to me where my logic is flawed? Thanks.
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  2. #2
    Senior Member
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    Your logic is quite right. You're just missing a step:

    $\displaystyle \binom{n-1}{r-k-1}=\frac{(n-1)!}{(r-k-1)!(n-1-r+k+1)!}$

    $\displaystyle =\frac{(n-1)!}{(r-k-1)!(n-r+k)!}$

    $\displaystyle =\frac{(n-1)!}{(n-1-n+r-k)!(n-r+k)!}$

    $\displaystyle =\binom{n-1}{n-r+k}$.
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  3. #3
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    @hatsoff
    Thank you very much that helps a lot. I'll have to remember to check things like that in the future.
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