Let $n$ and $k$ be integers with $1 \leq k \leq n$. Show that:
$$\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac12{2n+2 \choose n+1} - {2n \choose n}$$
I was told this is supposed to use a combinatorial proof and while I'm not that comfortable with that, many similar proofs use a mathematical proof that shows equality. Any guidance would be much appreciated.
The Combinatorial Proof goes as follows:
Consider three boxes $A$,$B$ and $C$.
Box $A$ contains $n$ balls, Box $B$ contains $n$ balls, and Box $C$ contains $2$ balls.
We want to choose $n+1$ balls out of these three boxes.
The total number of ways is $C(2n+2,n+1)$.
This can be split as follows.
You choose $2$ balls from box $C$ and the remaining $n-1$ balls from boxes $A$ and $B$ or you choose $1$ ball from box $C$ and the remaining $n$ balls from boxes $A$ and $B$ or you choose all the $n+1$ balls from boxes $A$ and $B$.
$\textbf{Case 1:}$
Suppose you choose $2$ balls from box $C$.
There is only one way of choosing these $2$ balls from the box $C$.
We now need the number of ways of choosing $n-1$ balls from boxes $A$ and $B$.
Say if you choose $k$ balls from box $A$, you need to choose $n-k-1$ balls from box $B$.
Number of ways of choosing $k$ balls from box $A$ and $n-k-1$ balls from box $B$ is given by $C(n,k)C(n,n-k-1)$.
Now summing over all possible values of $k$ from $0$ to $n-1$ we get,
Total number of ways is $\displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,n-k-1)$.
Now, $C(n,r) = C(n,n-r)$. (For a combinatorial argument of this see the bottom of the post).
Hence, Total number of ways for the current case is $1 \times \displaystyle \sum_{k=0}^{n-1} C(n,k)C(n,k+1) = \displaystyle \sum_{k=1}^{n} C(n,k-1)C(n,k)$.
$\textbf{Case 2:}$
Suppose you choose $1$ ball from the box $C$. You need to now choose the remaining $n$ balls from boxes $A$ and $B$.
Number of ways of choosing $1$ ball from box $C$ is given by $C(2,1) = 2$.
There are $2n$ balls in total in boxes $A$ and $B$ and hence the total number of ways of choosing $n$ balls from boxes $A$ and $B$ is $C(2n,n)$.
Hence, the total number of ways for the current case is $2 \times C(2n,n)$.
$\textbf{Case 3:}$
Suppose you choose all the $n+1$ balls from boxes $A$ and $B$.
Number of ways of choosing no balls from box $C$ is $1$.
We now need to count the number of ways of choosing $n+1$ balls from boxes $A$ and $B$.
This is similar to Case 1 and we get the total number of ways for this case is $1 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,n+1-k) = \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1)$.
(Note: Here $k$ goes from $1$ to $n$ since we need to choose atleast $1$ from one of the boxes and we can choose atmost $n$ from each box).
(Also Note: You could directly state that Case 1 and Case 3 should be the same since $C(2n,n-1)=C(2n,n+1)$)
Combine all the three cases to get
$2 \times \displaystyle \sum_{k=1}^{n} C(n,k)C(n,k-1) + 2 \times C(2n,n) = C(2n+2,n+1)$.
Now do the desired algebraic manipulation to get the result.
$\textbf{APPENDIX:}$
The combinatorial argument of $C(n,r) = C(n,n-r)$ is that the number of ways of selecting $r$ out of $n$ things is the same as the number of ways of discarding $n-r$ out of $n$ things.
Using standard binomial identities, this can be proven as follows: $$ \begin{align} \sum_{k=1}^n \binom{n}{k}\binom{n}{k-1}&=\sum_{k=1}^n \binom{n}{n-k}\binom{n}{k-1}\\ &=\binom{2n}{n-1}\\ &=\binom{2n+1}{n}-\binom{2n}{n}\\ &=\frac{n+1}{2n+2}\binom{2n+2}{n+1}-\binom{2n}{n}\\ &=\frac{1}{2}\binom{2n+2}{n+1}-\binom{2n}{n} \end{align} $$ Each identity can be given a simple combinatorial justification.
1. $$ \binom{n}{k}=\binom{n}{n-k} $$ The number of ways to choose $k$ items from $n$ is the number of ways to choose the complement of a set of $k$ items from $n$.
2. $$ \sum_{k=1}^n \binom{n}{n-k}\binom{n}{k-1}=\binom{2n}{n-1} $$ Given a set of green marbles, numbered $1...n$ and red marbles numbered $1...n$, count the choices of $n-1$ marbles from the aggregate $2n$ by counting the choices of $n-k$ green marbles and $k-1$ red marbles.
3. $$ \binom{2n}{n-1}+\binom{2n}{n}=\binom{2n+1}{n} $$ Given a set of green marbles numbered $1...2n$ and $1$ red marble, a choice $n$ marbles from the aggregate $2n+1$ either has the red marble, and thus $n-1$ of the $2n$ green marbles, or doesn't have the red marble, and thus $n$ of the $2n$ green marbles.
4. $$ (2n+2)\binom{2n+1}{n}=(n+1)\binom{2n+2}{n+1} $$ Putting $n+1$ marbles from a bag of $2n+2$ into a box and then choosing $1$ from the box is the same as choosing $1$ from the bag of $2n+2$ marbles and then putting $n$ of the remaining $2n+1$ marbles into the box.
Hint: Use $\displaystyle {n \choose k} = {n \choose n-k}$ and try to count something which equals $\displaystyle {n \choose n-k}{n \choose k-1}$
This identity can be interpreted as counting paths in $\Bbb Z^2$, each step being either a up $(+1,+1)$ or a down $(+1,-1)$.
The first remark is that $\binom{2n+2}{n+1}$ counts the total number of paths from $(0,0)$ to $(2n+2,0)$. Indeed, such a path is made of $n+1$ ups and $n+1$ downs, and we need only specify which steps among the $2n+2$ are ups. Similarly $\binom{2n}{n}$ is the number of paths from $(2,0)$ to $(2n+2,0)$ and it is now easy to see that $\frac12{2n+2 \choose n+1} - {2n \choose n}$ is actually the number of paths from $(0,0)$ to $(2n+2,0)$ which start with two up steps.
Since the fisrt two steps are fixed, we need only consider the last $2n$ steps between $(2,2)$ and $(2n+2,0)$. The number of paths with $k$ downs in the first $n$ steps and $k'$ ups in the last $n$ steps is $\binom{n}{k}\binom{n}{k'}$. Moreover, the numbers $k$ and $k'$ are constrained by the condition to reach $(2n+2,0)$, which is $2 + (n-k) - k -(n-k') + k' = 0$ or equivalently $k'=k-1$. Summing over all possible values for $k$ we obtain the desired formula $$ \frac12{2n+2 \choose n+1} - {2n \choose n} = \sum_{k=1}^n \binom{n}{k}\binom{n}{k-1}. $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\color{#66f}{\large\sum_{k = 1}^{n}{n \choose k}{n \choose k - 1}} =\sum_{k = 1}^{n}{n \choose k}\ \overbrace{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{k}} \,{\dd z \over 2\pi\ic}}^{\ds{n \choose k - 1}} \\[3mm]&=\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n} \sum_{k = 1}^{n}{n \choose k}\pars{1 \over z}^{k}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n} \bracks{\pars{1 + {1 \over z}}^{n} - 1}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n}\, \bracks{{\pars{1 + z}^{n} \over z^{n}} - 1}\,{\dd z \over 2\pi\ic} \\[3mm]&=\underbrace{\oint_{\verts{z}\ =\ 1}% {\pars{1 + z}^{2n} \over z^{n}}\,{\dd z \over 2\pi\ic}}_{\ds{2n \choose n - 1}}\ -\ \underbrace{\oint_{\verts{z}\ =\ 1}\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic}} _{\ds{=\ 0}} ={2n \choose n - 1} \\[5mm]&=\color{#66f}{\large\half\,{2n + 2 \choose n + 1} - {2n \choose n}} \end{align}
Note that \begin{align}\color{#c00000}{\half\,{2n + 2 \choose n + 1} - {2n \choose n}}&= \half\,{\pars{2n + 2}! \over \pars{n + 1}!\pars{n + 1}!} -{\pars{2n}! \over n!\,n!} ={\pars{2n + 2}! - 2\pars{n + 1}^{2}\pars{2n}!\over 2\bracks{\pars{n + 1}!}^{2}} \\[3mm]&={2\pars{n + 1}\pars{2n + 1}\pars{2n}! - 2\pars{n + 1}^{2}\pars{2n}!\over 2\pars{n + 1}n\pars{n - 1}!\pars{n + 1}!} \\[3mm]&={\pars{n + 1}\pars{2n + 1} - \pars{n + 1}^{2} \over \pars{n + 1}n}\, {\pars{2n}! \over \pars{n - 1}!\pars{n + 1}!} \\[3mm]&={\pars{n + 1}\bracks{\pars{2n + 1} - \pars{n + 1}} \over \pars{n + 1}n}\, {2n \choose n - 1} = \color{#c00000}{{2n \choose n - 1}} \end{align}
Both sides count the number of ways of choosing $n+1$ items from the set $\{1,2,3,\ldots,2n\}$.
The $k^\text{th}$ term in the sum on the left counts selections where $k$ items are chosen from the first half of the list, i.e. from $\{1,2,\ldots,n\}$ and $n+1-k$ items are chosen from the second half of the list, i.e. from $\{n+1,n+2,\ldots,2n\}$ (which means that $k-1$ items from the second half are not chosen). By the pigeonhole principle, at least one item must come from the first half, which explains why the sum starts with $k=1$.
We are choosing approximately half of the items in the set. We could make it exactly half by expanding the set we are selecting from to include $0$ and $2n+1$. To obtain the desired count we would then exclude those selections in which either $0$ or $2n+1$ (or both) were among the chosen. Expanding the set is nice since, with $n+1$ items chosen, there are also $n+1$ items unchosen, and we may partition the set of selections into pairs, each consisting of a selection and its complement. (The complement of a selection is the selection where the designations "chosen" and "unchosen" are interchanged). In each pair, one selection will include $0$ and the other will not. So we may eliminate the selections containing $0$ simply by taking the selection from each pair that does not contain $0$ and discarding the one that does. This explains the factor of $\frac{1}{2}$ in the first term. From those selections that remain, we now subtract those that contain $2n+1$ (and hence $n$ items from $\{1,2,\ldots,2n\}$.
