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Recently I have taken a look at this vandermonde identity proof Inductive Proof for Vandermonde's Identity?. \begin{align*} \binom{m + (n+1)}r &= \binom{m+n}r + \binom{m+n}{r-1}\\ &= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\ &= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\ &= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\ &= \sum_{k=0}^r \binom mk \binom{n+1}{r-k} \end{align*}

I am confused mostly at this part \begin{align*} &= \sum_{k=0}^r \binom mk\binom n{r-k} + \sum_{k=0}^{r-1} \binom mk\binom{n}{r-1-k}\\ &= \binom mr + \sum_{k=0}^{r-1} \binom mk\biggl(\binom n{r-k} + \binom n{r-1-k}\biggr)\\ \end{align*}

How did this \begin{align*} &= \sum_{k=0}^r \binom mk\binom n{r-k} \end{align*} turn into \begin{align*} &= \binom mr \end{align*}

which identity did this person use? I am such that it could magically become C(m,r)? I am also confused on how \begin{align*} \binom{n+1}0 \end{align*} appears in here \begin{align*} &= \binom mr\binom{n+1}0 + \sum_{k=0}^{r-1} \binom mk\binom{n+1}{r-k}\\ \end{align*} If someone could help that would be very helpful, I didn't comment on the post due to it being 8 years, posted already. If someone could help and point where I should study that would be very helpful. Thank you!

RobPratt
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Dwikavindra Haryo Radithya
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1 Answers1

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Do you mean this step?

$\displaystyle\sum_{k=0}^r {m \choose k}{n \choose r-k} = {m \choose r} + \sum_{k=0}^{r-1}{m\choose k}{n\choose r-k}$

Also for your second question, as $0! \equiv 1$ we'll have ${n+1\choose 0} = 1$.

bartholovidius
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