Proving $\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$

Question

Prove the identity: $\displaystyle\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$

It looks a bit similar to the "no gets their own hat back" problem or inclusion exclusion or non distinct balls in bins.

Trying to find a combinatorial solution seems like impossible because of the alternating sum (how can we explain inclusion exclusion?).

Trying to expand the RHS doesn't help nor using any of the simple identities I know of (like Pascal's).

Any hints or directions please?

Note: no integrals, no generating functions nor use of other identities without proving them.

Edit: I think I got it:

LHS:

n non distinct balls to r bins such that every bin has at least one ball, spread 1 ball to each bin, we're left with n-r balls to r bins.

RHS:

General case: $\binom {n+r-1}{r-1}$

complement: at least one bin is empty; 1 bin is empty, choose that bin $\binom r 1$ and spread the balls: $\binom{n+r-1-1}{r-1-1}$, do this up to r empty bins.

Since we have many over counting, we'll apply the inclusion exclusion principle and we got what we desired.

Answer 1

$$\begin{align} \sum_{k=0}^r(-1)^k\binom rk\binom{n+r-k-1}{r-k-1} &=\sum_{k=0}^r(-1)^k\binom rk\binom{-n-1}{r-k-1}(-1)^{r-k-1}&&(1)\\ &=(-1)^{r-1}\sum_{k=0}^r\binom rk\binom{-n-1}{r-1-k}\\ &=(_1)^{r-1}\binom{r-n-1}{r-1}&&(2)\\ &=(-1)^{r-1}\binom{n-1}{r-1}(-1)^{r-1}&&(3)\\ &=\binom{n-1}{r-1}\qquad\blacksquare \end{align}$$

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(1): using Upper Negation
(2): using Vandermonde Identity
(3): using Upper Negation
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$\color{gray}{\text{Proof of Upper Negation}}$ $$\color{gray}{\begin{align}\\ \binom ab&=\frac{a^{\underline{b}}}{b!} =\frac{a(a-1)(a-2)\cdots(a-b+1)}{b!}\\ &=(-1)^b \frac{[-a][-(a-1)][-(a-2)]\cdots[-(a-b+1)]}{b!}\\ &=(-1)^b \frac{(b-1-a)\cdots(2-a)(1-a)(-a)}{b!}\\ &=(-1)^b \binom{b-a-1}b\end{align}}$$

$\color{gray}{\text{Proof of Vandermonde Identity}}$ $$\color{gray}{\begin{align}\\ \sum_{i=0}^a\binom ai x^i\sum_{j=0}^b\binom bjx^j &=(1+x)^a(1+x)^b=(1+x)^{a+b}\\ [x^n]: \sum_{i+j=n}\binom ai\binom bj &=\underbrace{\sum_{i=0}^n \binom ai\binom b{n-i}=\binom{a+b}n}_{\text{Vandermonde Identity}} \end{align}}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{r}\pars{-1}^{k}{r \choose k}{n + r - k - 1 \choose r - k - 1}} = \sum_{k = 0}^{r}\pars{-1}^{k}{r \choose k}{-n - 1 \choose r - k - 1} \pars{-1}^{r - k - 1} \\[5mm] = &\ \pars{-1}^{r + 1}\sum_{k = 0}^{r}{r \choose k} \bracks{z^{r - k - 1}}\pars{1 + z}^{-n - 1} = \pars{-1}^{r + 1}\bracks{z^{r - 1}}\pars{1 + z}^{-n - 1} \sum_{k = 0}^{r}{r \choose k}z^{k} \\[5mm] = &\ \pars{-1}^{r + 1}\bracks{z^{r - 1}}\pars{1 + z}^{-n - 1 + r} = \pars{-1}^{r + 1}{-n - 1 + r \choose r - 1} \\[5mm] = &\ \pars{-1}^{r + 1}{n - 1 \choose r - 1}\pars{-1}^{r - 1} = \bbx{\large{n - 1 \choose r - 1}} \\ & \end{align}

Answer 3

We seek to evaluate

$$\sum_{k=0}^r (-1)^k {r\choose k} {n+r-k-1\choose r-k-1} = \sum_{k=0}^r (-1)^k {r\choose k} {n+r-k-1\choose n}.$$

Note that the second binomial coefficient is zero when $k=r$ because $(n-1)^\underline{n} = 0.$ Continuing we find

$$[z^n] (1+z)^{n+r-1} \sum_{k=0}^r (-1)^k {r\choose k} (1+z)^{-k} \\ = [z^n] (1+z)^{n+r-1} \left(1-\frac{1}{1+z}\right)^r = [z^n] (1+z)^{n+r-1} \frac{z^r}{(1+z)^r} \\ = [z^{n-r}] (1+z)^{n-1} = {n-1\choose n-r} = {n-1\choose r-1}.$$

Answer 4

Actually, something very cute is that i've found this result alone. I was analazying a problem, then i realized that in $$ i_1 + i_2 + i_3 +...+i_k=n$$ if i have some integers that are equal to 0, i can "delete" such integer from the Diophante equation, let's suppose that you have i_1=0, then this is equal to $$i_2+i_3+...i_k$$, there is a bijection. Now, whe can say that if we are interested to find this result: $$\binom{n+k-1}{k-1} - \binom{n-1}{k-1}= x$$ We are interested in individuading all the configuration that have atleast one integer that is 0. Now this procidure needs the PIE. Firstly, you should try to exclude one integer, doing so you can calculate all the configurations such that integer k equals to 0. But when you calculate $$\binom{n+k-2}{k-2}$$ you are actually counting the configurations where other integers are identical to 0. Now, thanks to this, we can say that $$\binom{n+k-1}{k-1} - \binom{n-1}{k-1}$$ needs PIE to be calculated. Now let's try to solve the first step, let's try to set the case where only one integer is 0. Now, we can select the integer directly from the Diophante equation in $$ \binom{k}{1}$$ ways, but in PIE that 1 is actually a variable q that in this case goes from 1 to k-1. Why k-1? Well, because if you have all the integers identically to 0, the Diophante equation cannot have any result rather then 0! So, you'll end up with: $$\binom{n+k-1}{k-1} -\binom{n-1}{k-1}= \sum_{q=1}^{k-1}\binom{k}{q} \binom{n+k-q-1}{k-q-1} (-1)^{q-1} $$ But if we try to manipulate more the thing: $$-\binom{n-1}{k-1}=-\binom{n+k-1}{k-1}+\sum_{q=1}^{k-1}\binom{n+k-q-1}{k-q-1}$$s Finally: $$\binom{n-1}{k-1}= \binom{n+k-1}{k-1} + \sum_{q=0}^{k-1} \binom{k}{q} \binom{n+k-q-1}{k-q-1}(-1)^q$$. But if you look closely, you can complete the sigma: $$\binom{n-1}{k-1}= \sum_{q=0}^{k-1} \binom{k}{q} \binom{n+k-q-1}{k-q-1} (-1)^q$$ Because $$\binom{n+k-1}{k-1}= \binom{n+k-0-1}{k-0-1}$$