Proof of $\sum_{m=0}^{n}\binom{m}{j}\binom{n-m}{k-j}=\binom{n+1}{k+1}$ (Another form of the Chu–Vandermonde identity)

Question

I found an identity $$ \sum_{m = 0}^{n}\binom{m}{j}\binom{n - m}{k - j} = \binom{n + 1}{k + 1} $$

from Sum of Binomial Coefficients.

• I can prove this in a combinatorial way: from $n + 1$ balls, choose one ball as a boundary and pick up $j$ balls from the left and $k - j$ balls from the right, for every possible boundary balls.
• And then I can prove this is identical to choosing $k+1$ balls from $n+1$ balls.
• On the other hand, the wiki says that you can prove this by using negative binomial series expansion. I have tried few attempts but wasn't very successful.

Could anyone share a proof using binomial series expansion ?. Thank you.

Answer 1

Recall that $$ (1-x)^{-s} = \sum_{t=0}^{\infty} \binom{s+t-1}{t} x^t . $$ Recall also the notation that for a (formal) power series $f$, $[x^a]f$ denotes the coefficient of $x^a$ in $f$. In particular $[x^t](1-x)^{-s} = \binom{s+t-1}{t}$. And specifically, $$ [x^{m-j}](1-x)^{-(j+1)} = \binom{m-j+j}{m-j} = \binom{m}{j} $$ and $$ [x^{n-m-k+j}](1-x)^{-(k-j+1)} = \binom{n-m}{k-j} . $$ Multiplying these and summing over $m$ gives the coefficient of $$ x^{(m-j) + (n-m-k+j)} = x^{n-k} $$ in $(1-x)^{-(k+2)}$, thus $$ \begin{multline*} \sum_{m=0}^{\infty} \binom{m}{j}\binom{n-m}{k-j} = [x^{n-k}](1-x)^{-(k+2)} = \binom{(k+2)+(n-k)-1}{n-k} \\= \binom{n+1}{n-k} = \binom{n+1}{k+1}. \end{multline*} $$ The limits of summation for $m$ don't matter too much; the terms are only nonzero for $j \leq m \leq n-k+j$ anyway, and $n-k+j \leq n$, so it doesn't make a difference whether we sum up to $n$ or up to $\infty$.

Answer 2

Given that $$ \frac{x^j}{(1 - x)^{j+1}} = \sum_{m = 0}^\infty \binom{m}{j}x^m, $$

We have \begin{align} \sum_{m_1 + m_2 = n} \binom{m_1}{j} \binom{m_2}{k - j} &= [x^n] \frac{x^j}{(1 - x)^{j + 1}} \frac{x^{k - j}}{(1 - x)^{k - j + 1}} \\ &= [x^n] \frac{x^k}{(1 - x)^{k + 2}} \\ &= [x^{n + 1}] \frac{x^{k + 1}}{(1 - x)^{k + 2}} \\ &= \binom{n + 1}{k + 1}. \end{align}

Answer 3

thanks to How to apply Vandermonde's Identity regarding summation bounds ? $\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}=\binom{-k-j-2}{n-j-k}$ for explaining the technique of changing summation bounds

proof of $\binom{-n}{k} =(-1)^{k}\binom{n+k-1}{k}$ What does $\binom{-n}{k}$ mean?

proof of $\sum _{m=0}^{k}\binom{a}{m}\binom{b}{k-m} =\binom{a+b}{k}$ https://en.wikipedia.org/wiki/Vandermonde%27s_identity#Algebraic_proof

$$prove\ \sum _{m=0}^{n}\binom{m}{j}\binom{n-m}{k-j} =\binom{n+1}{k+1}$$ $$\sum _{m=0}^{n}\binom{m}{j}\binom{n-m}{k-j}$$ $$USE:\ \binom{m}{j} =\binom{m}{m-j}$$ $$\sum _{m=0}^{n}\binom{m}{m-j}\binom{n-m}{n-m-k+j}$$ $$USE:\binom{-n}{k} =(-1)^{k}\binom{n+k-1}{k}$$ $$\sum _{m=0}^{n}\binom{-( -m)}{m-j}\binom{-( -n+m)}{n-m-k+j}$$ $$\sum _{m=0}^{n} (-1)^{m-j}\binom{( -m) +( m-j) -1}{m-j} (-1)^{n-m-k+j}\binom{( -n+m) +( n-m-k+j) -1}{n-m-k+j}$$ $$\sum _{m=0}^{n} (-1)^{m-j}\binom{-j-1}{m-j} (-1)^{n-m-k+j}\binom{-k+j-1}{n-m-k+j}$$ $$\sum _{m=0}^{n} (-1)^{m-j} (-1)^{n-m-k+j}\binom{-j-1}{m-j}\binom{-k+j-1}{n-m-k+j}$$ $$\sum _{m=0}^{n} (-1)^{n-k}\binom{-j-1}{m-j}\binom{-k+j-1}{n-m-k+j}$$ $$(-1)^{n-k}\sum _{m=0}^{n}\binom{-j-1}{m-j}\binom{-k+j-1}{n-m-k+j}$$ $$USE:k< 0\Longrightarrow \binom{n}{k} =0$$ $$m-j< 0$$ $$m< j\Longrightarrow \binom{-j-1}{m-j} =0$$ $$n-m-k+j< 0$$ $$n-k+j< m\Longrightarrow \binom{-k+j-1}{n-m-k+j} =0$$ $$indices\ that\ give\ non-zero:j\leq m\leq n-k+j$$ $$change\ the\ summation\ bounds$$ $$(-1)^{n-k}\sum _{m=j}^{n-k+j}\binom{-j-1}{m-j}\binom{-k+j-1}{n-m-k+j}$$ $$shift\ the\ summation\ by\ p=-j,\ USE:\sum _{n=s}^{t} f( n) =\sum _{n=s+p}^{t+p} f( n-p)$$ $$(-1)^{n-k}\sum _{m=0}^{n-k}\binom{-j-1}{( m+j) -j}\binom{-k+j-1}{n-( m+j) -k+j}$$ $$(-1)^{n-k}\sum _{m=0}^{n-k}\binom{-j-1}{m}\binom{-k+j-1}{n-m-k}$$ $$Vandermonde's\sum _{m=0}^{k}\binom{a}{m}\binom{b}{k-m} =\binom{a+b}{k}$$ $$(-1)^{n-k}\sum _{m=0}^{n-k}\binom{-j-1}{m}\binom{-k+j-1}{( n-k) -m}$$ $$(-1)^{n-k}\binom{( -j-1) +( -k+j-1)}{( n-k)}$$ $$(-1)^{n-k}\binom{( -k-1) -1}{( n-k)}$$ $$(-1)^{n-k}\binom{( -k-1) -( n-k) +( n-k) -1}{( n-k)}$$ $$(-1)^{n-k}\binom{( -1-n) +( n-k) -1}{( n-k)}$$ $$USE:\binom{-n}{k} =(-1)^{k}\binom{n+k-1}{k}$$ $$\binom{n+1}{n-k}$$ $$USE:\ \binom{m}{j} =\binom{m}{m-j}$$ $$\binom{n+1}{( n+1) -( n-k)}$$ $$\binom{n+1}{k+1}$$

I wasn't able to prove it using $(1 + x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k}$, and I would like to see a proof that doesn't use the coefficient of operator, just for proof of concept.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{m = 0}^{n}{m \choose j} {n - m \choose k - j}} = \bracks{z^{n}}\sum_{\ell = 0}^{\infty}z^{\ell} \sum_{m = 0}^{\ell}{m \choose j}{\ell - m \choose k - j} \\[5mm] = & \ \bracks{z^{n}}\sum_{m = 0}^{\infty} {m \choose j}\sum_{\ell = m}^{\infty}z^{\ell} {\ell - m \choose k - j} \\[5mm] = & \ \bracks{z^{n}}\sum_{m = 0}^{\infty} {m \choose j}\sum_{\ell = 0}^{\infty}z^{\ell + m}\, {\ell \choose k - j} \\[5mm] = & \ \bracks{z^{n}}\bracks{\sum_{m = 0}^{\infty} {m \choose j}z^{m}} \bracks{\sum_{\ell = 0}^{\infty} {\ell \choose k - j}z^{\ell}}\label{1}\tag{1} \\ - &------------------------------ \\ & \mbox{However,}\ \color{red}{\sum_{k = 0}^{\infty}{k \choose s}z^{k}} = \sum_{k = s}^{\infty}{k \choose s}z^{k} = \sum_{k = 0}^{\infty}{k + s \choose s}z^{k + s} \\[5mm] = & \ z^{s}\sum_{k = 0}^{\infty}{k + s \choose k}z^{k} = z^{s}\sum_{k = 0}^{\infty}\bracks{{-s - 1 \choose k}\pars{-1}^{k}}z^{k} \\[5mm] = & \ \color{red}{z^{s}\pars{1 - z}^{-s - 1}}\label{2}\tag{2} \\ - & ------------------------------ \\ & \mbox{Therefore,}\ (\ref{1})\ \mbox{and}\ (\ref{2}) \implies \\ & \color{#44f}{\sum_{m = 0}^{n}{m \choose j} {n - m \choose k - j}} \\[5mm] = & \ \bracks{z^{n}}\bracks{z^{j}\,\pars{1 - z}^{-j - 1}\,} \bracks{z^{k - j}\,\pars{1 - z}^{-k + j - 1}\,} \\[5mm] = & \ \bracks{z^{n}}z^{k}\,\,\pars{1 - z}^{\pars{-j - 1}\ +\ \pars{-k + j - 1}}\,\,\,\,\, = \bracks{z^{n - k}}\pars{1 - z}^{-k - 2} \\[5mm] = & \ {-k - 2 \choose n - k}\pars{-1}^{n - k} \\[5mm] = & \ {-\bracks{-k - 2} + \bracks{n - k} - 1 \choose n - k} = {n + 1 \choose n - k} \\[5mm] = & \ \bbx{\color{#44f}{n + 1 \choose k + 1}} \\ & \end{align}

Answer 5

Wiki says that you can prove this by using negative binomial series expansion.

Here is the way:

$$\sum_{m=0}^{n}\binom{m}{j}\binom{n-m}{k-j}=\sum_{m=j}^{n}\binom{m}{m-j}\binom{n-m}{n-m-k+j}$$$$=\sum_{m=j}^{n}\left(-1\right)^{m-j}\binom{-j-1}{m-j}\left(-1\right)^{n-m-k+j}\binom{-k+j-1}{n-m-k+j}$$ Setting $m-j \mapsto m$ yields:

$$=\left(-1\right)^{n-k}\sum_{m=0}^{n-j}\binom{-j-1}{m}\binom{-k+j-1}{n-m-k}$$$$=\left(-1\right)^{n-k}\binom{-k-2}{n-k}$$$$=\binom{n+1}{n-k}=\binom{n+1}{k+1}$$

So we showed that :

$$\bbox[5px,border:2px solid #00A000]{\sum_{m=0}^{n}\binom{m}{j}\binom{n-m}{k-j}=\binom{n+1}{k+1}}$$

Which is the claim.

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Note:

Here Vandermonde's convolution is generalized:

$$\large\sum_{k=0}^{\large\sum_{\mu=1}^{i}n_\mu}\binom{\large\sum_{\mu=1}^{i}n_\mu}{k}x^{k}=\left(1+x\right)^{\large\sum_{\mu=1}^{i}n_\mu}$$$$=\large\prod_{\mu=1}^{ i}\left(1+x\right)^{n_\mu}=\prod_{\mu=1}^{ i}\sum_{k_{\mu }=0}^{n_{\mu }}\binom{n_\mu}{k_\mu}x^{k_\mu}$$$$=\large{\sum_{k=0}^{\large\sum_{\mu=1}^{i}n_\mu }\sum_{\sum_{\mu=1}^{i} v_\mu=k}^{ }}\;\;{\prod_{\mu=1}^{ i}}a_{v_\mu}\large x^{k}$$

Where: $$\large a_{v_\mu}=\binom{n_\mu}{v_\mu}$$

Comparing the coefficient of $x^k$ on the both sides follows:

$$\bbox[5px,border:2px solid #00A000]{\binom{\large\sum_{\mu=1}^{i}n_\mu}{k}=\large\sum_{\sum_{\mu=1}^{i}v_{\mu}=k}^{ }\;\;\prod_{\mu=1}^{i}\large \binom{n_\mu}{v_\mu}}$$

Regular Vandermonde's convolution is a special case of this relation which can be derived by setting $i \mapsto 2$