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How to show that

$$\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$$

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3 Answers 3

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Here is an answer based upon Short Proofs of Saalschütz's and Dixon's theorems by Ira Gessel and Dennis Stanton.

The answer is provided in three steps

  • Step 1: Relationship between coefficients of the constant term of a bivariate Laurent series and a transformed of it.

  • Step 2: Application of this relationship to a specific Laurent series with the coefficient of the constant term being strongly related with OPs binomial identity.

  • Step 3: Substitutions to reveal the symmetry of the binomial identity and to transform it finally to OPs stated binomial identity.

We consider formal Laurent series \begin{align*} A(x,y)=\sum_{m,l}a_{m,l}x^my^l \end{align*} in which for only finitely many pairs $(m,l)$ the coefficients $a_{m,l}\ne 0$ with $m<0$ or $l<0$. Let $[x^n]$ denote the coefficient of $x^n$ in a series. We start with a useful observation

Step 1: Let $A(x,y)$ be a formal Laurent series. The following is valid \begin{align*} [x^0y^0]A\left(\frac{x}{1+y},\frac{y}{1+x}\right)=[x^0y^0]\frac{1}{1-xy}A(x,y)\tag{1} \end{align*}

By linearity it is sufficient to prove (1) by considering $A(x,y)=x^ly^m$ with $l,m\in \mathbb{Z}$. We have to show \begin{align*} [x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}=[x^0y^0]\frac{x^ly^m}{1-xy}\tag{2} \end{align*}

We show the LHS of (2) fulfills \begin{align*} [x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}= \begin{cases} 1&\qquad\text{if }l=m\leq 0\\ 0&\qquad\text{otherwise} \end{cases}\tag{3} \end{align*}

  • If $l>0$ or $m>0$ the result in (3) is $0$ since the binomial expansion of $(1+y)^{-l}(1+x)^{-m}$ has no negative powers.

  • If $l=-r,m=-s$ with $r,s\geq 0$ we note that \begin{align*} [x^0y^0]\frac{(1+x)^s(1+y)^r}{x^ry^s}&=[x^ry^s](1+x)^s(1+y)^r\\ &=\binom{s}{r}\binom{r}{s}= \begin{cases} 1&\qquad\text{if }r=s\\ 0&\qquad\text{if }r\neq s \end{cases} \end{align*}

and (3) follows.

We look at the RHS of (2) and observe \begin{align*} [x^0y^0]\frac{x^ly^m}{1-xy}=[x^{-l}y^{-m}]\sum_{j\geq 0}(xy)^j =\begin{cases} 1&\qquad\text{if }l=m\leq 0\\ 0&\qquad\text{otherwise} \end{cases} \end{align*} which is the same as (3) and the claim (2) follows.

Step 2: We now apply (1) to the Laurent series \begin{align*} f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^n}\qquad\qquad\qquad l,m\in\mathbb{Z},n\in\mathbb{N} \end{align*}

We obtain

\begin{align*} &f\left(\frac{x}{1+y},\frac{y}{1+x}\right)\\ &\qquad=\left(\frac{x}{1+y}-\frac{y}{1+x}\right)^n \cdot\frac{(1+y)^l}{x^l}\cdot\frac{(1+x)^m}{y^m}\cdot\left(1-\frac{xy}{(1+x)(1+y)}\right)^{-n}\\ &\qquad=\frac{\left((x-y)+(x^2-y^2)\right)^n}{(1+x)^n(1+y)^n}\cdot \frac{(1+y)^l}{x^l}\frac{(1+x)^m}{y^m}\cdot \frac{(1+x)^n(1+y)^n}{(1+x+y)^n}\\ &\qquad=\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\tag{4} \end{align*}

on the other hand we get

\begin{align*} \frac{1}{1-xy}f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{5} \end{align*}

equating (4) and (5) we obtain according to (2) \begin{align*} [x^0y^0]\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}=[x^0y^0]\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{6} \end{align*}

The LHS of (6) gives \begin{align*} [x^0y^0]&\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\\ &=[x^ly^m]\sum_{k}\binom{n}{k}(-y)^kx^{n-k}(1+x)^m(1+y)^l\\ &=\sum_{k}\binom{n}{k}(-1)^k[x^{l-n+k}y^{m-k}] \sum_{i=0}^m\binom{m}{i}x^i\sum_{j=0}^l\binom{l}{j}y^j\\ &=\sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k} \end{align*}

In order to consider the RHS of (6), we do a little trick. We set $x=xz$ and $y=yz$ and note that \begin{align*} [x^0y^0]f(x,y)&=[x^0y^0z^0]f(xz,yz)\\ &=[x^0y^0z^0]\frac{(xz-yz)^n}{(xz)^l(yz)^m(1-z^2xy)^{n+1}}\\ &=[x^0y^0z^0]\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\tag{7} \end{align*} We observe that (7) is zero if $l+m-n$ is odd. If $l+m-n=2r$ is even, we obtain \begin{align*} [x^0y^0z^0]&\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\\ &=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{-(n+1)}{j}(-z^2xy)^j\\ &=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{n+j}{j}(z^2xy)^j\\ &=\binom{n+r}{r}[x^{l-r}y^{m-r}]\sum_{j=0}^nx^k(-y)^{n-k}\\ &=\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{8} \end{align*}

With focus on (8) at which $l+m-n=2r$ is even we conclude

\begin{align*} \sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k} =\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{9} \end{align*}

Step 3: We can now put more symmetry into (9) by setting

\begin{align*} n&=a+b\\ m&=c+a\\ l&=b+c \end{align*} We obtain \begin{align*} \sum_{k}(-1)^k\binom{a+b}{k}\binom{c+a}{c-a+k}\binom{b+c}{b-a+k}=(-1)^{a}\binom{a+b+c}{a+b}\binom{a+b}{a} \end{align*} Finally multiplying this identity with $(-1)^a$ and shifting the index $k\rightarrow k+a$ we obtain \begin{align*} \sum_{k}(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}&=\binom{a+b+c}{a+b}\binom{a+b}{a}\\ &=\frac{(a+b+c)!}{a!b!c!} \end{align*} and the claim follows.

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Dixon's identity is a notable consequence of MacMahon's master theorem (MMT). The proof of Dixon's identity from MMT is fairly concise and elegant, but the proof of MMT requires some non-trivial linear algebraic concepts, in particular symmetric powers of linear maps and vector spaces.

MMT: Let $\mathbf A = [a_{ij}]_{n\times n}$, $\mathbf X = \operatorname{diag}(x_1,\dots,x_n)$ and $\mathbf k = (k_1,\dots,k_n) \geq 0$. Then, $$ [\mathbf t^\mathbf k] \prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n a_{ij} t_j\right)^{k_i} = [\mathbf x^\mathbf k] \det(\mathbf I-\mathbf X \mathbf A)^{-1}. $$


Dixon's identity: Let $a,b,c \geq 0$, then $$ \sum\limits_{k} (-1)^k \binom{a+b}{a+k} \binom{b+c}{b+k} \binom{c+a}{c+k} = \binom{a+b+c}{a,b,c}. $$ Proof: We can represent the LHS as $$ (-1)^{a+b+c}[x^{b+c} y^{c+a} z^{a+b}] (x-y)^{a+b}(y-z)^{b+c}(z-x)^{c+a}, $$ as it expands into $$ \sum\limits_{k} \binom{a+b}{a+k}x^{b-k} (-y)^{a+k} \binom{b+c}{b+k} y^{c-k} (-z)^{b+k} \binom{c+a}{c+k} z^{a-k} (-x)^{c+k}. $$ Applying the MacMahon's theorem for $\mathbf{x} = (x,y,z)$ and $\mathbf{k}=(b+c,c+a,a+b)$, we get

$$ [x^{b+c} y^{c+a} z^{a+b}] (x-y)^{a+b}(y-z)^{b+c}(z-x)^{c+a} = [x^{b+c} y^{c+a} z^{a+b}] \begin{vmatrix} 1 & x & -x \\ -y & 1 & y \\ z & -z & 1 \end{vmatrix}^{-1}, $$ which rewrites into $$ [(xy)^c (xz)^b (yz)^a] \frac{1}{1+xy+yz+xz} = (-1)^{a+b+c}\binom{a+b+c}{a,b,c}. \square $$


MMT proof: Sum up LHS and RHS over $\mathbf k \geq 0$, both multiplied by $\mathbf x^\mathbf k$:

$$ \sum\limits_{\mathbf k \geq 0} [\mathbf t^\mathbf k]\prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n a_{ij} x_i t_j\right)^{k_i} = \det(\mathbf I - \mathbf X \mathbf A)^{-1}. $$

If we now substitute $\mathbf B = [a_{ij} x_i]_{n\times n} = \mathbf X \mathbf A$, the identity turns into

$$ \sum\limits_{\mathbf k \geq 0} [\mathbf t^\mathbf k]\prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n b_{ij} t_j\right)^{k_i} = \det(\mathbf I - \mathbf B)^{-1}. $$ This is an equivalent formulation of MMT that doesn't need any variables in RHS anymore. This allows us to rewrite it in better terms. Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of $\mathbf B$, then the RHS rewrites as $$ \det(\mathbf I - \mathbf B)^{-1} = \prod\limits_{i=1}^n \frac{1}{1-\lambda_i} = \prod\limits_{i=1}^n \sum\limits_{k=0}^\infty \lambda_i^k = \sum\limits_{\mathbf k \geq 0} \lambda^\mathbf k. $$

To advance here, we should rely on some properties of the $k$-th symmetric power $\operatorname{S}^k(\mathbf B)$ of $\mathbf B$. Let $(e_1,\dots,e_n)$ be the basis formed of eigenvectors of $\mathbf B$. Then, we can also construct a basis in the symmetric power of the underlying vector space, with basis vectors being eigenvectors of $\operatorname{S}^k(\mathbf B)$.

For this, we index eigenvalues and eigenvectors by tuples $\mathbf k = (k_1,\dots,k_n) \geq 0$ such that $k_1+\dots+k_n=k$. The tuple corresponds to an eigenvalue $\lambda^\mathbf k = \lambda_{1}^{k_1} \dots \lambda_{n}^{k_n}$ with an eigenvector

$$ e_{\mathbf k} = \sum\limits_{(j_1,\dots,j_k)} e_{j_1} \otimes \dots \otimes e_{j_k}, $$ where $(j_1,\dots,j_k)$ covers all possible orderings, in which each $j$ is included exactly $k_j$ times. Grouping products by $k$, and noting that the sum of eigenvalues is the trace of the linear map, we get

$$ \det(\mathbf I - \mathbf B)^{-1} = \sum\limits_{k=0}^\infty \operatorname{tr}\operatorname{S}^k(\mathbf B). $$

On the other hand, we can also express $\operatorname{tr}\operatorname{S}^k(\mathbf B)$ in coordinate form. The construction above to make a basis in the symmetric power of the underlying vector space works not only for eigenvalues of $\mathbf B$, but for any basis $(e_1,\dots,e_n)$ of the initial space. Now, we can find the trace as

$$ \operatorname{tr} \operatorname{S}^k(\mathbf B) = \sum\limits_{\substack{\mathbf k}} \frac{e_\mathbf k \cdot \operatorname{S^k}(\mathbf B) e_\mathbf k}{e_\mathbf k \cdot e_\mathbf k}, $$

where $\mathbf k$ covers all tuples $(k_1,\dots,k_n) \geq 0$ with $k_1+\dots+k_n = k$. Then, it rewrites into

$$ \sum\limits_\mathbf k \sum\limits_{(j_1,\dots,j_k)} \prod\limits_{s=1}^k b_{i_s j_s}, $$ where $(j_1,\dots,j_k)$ cover all orderings, in which each $j$ occurs exactly $k_j$ times, and $(i_1,\dots,i_k)$ is the lexicographically smallest such ordering. Thus, we sum up the products of $b_{i_s j_s}$ over all pairings $(i_s, j_s)$, in which every number $j$ occurs $k_j$ times as the second element. But the expression in the LHS of MMT,

$$ \sum\limits_{\mathbf k} [\mathbf t^\mathbf k]\prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n b_{ij} t_j\right)^{k_i} $$ actually counts the same thing if we limit summation to $k_1+\dots+k_n=k$. Indeed, for each $i$ we select $k_i$ elements that would be second in pairs when we expand the corresponding sum, raised to power $k_i$. At the same time, the variables $t_j$ keep track of how much each number occurred as the second pair element, making sure it's $k_j$ for each $j$. $\square$


As a final note, MMT itself is a partial (linear) version of an even more generic result, known as multivariate Lagrange implicit function theorem (LIFT).

Multivariate LIFT: Let $\mathbf x = (x_1,\dots,x_n)$ and let $R_1(\mathbf x), \dots, R_n(\mathbf x) \in \mathbf K[[\mathbf x]]$ be such that $$ R_j = x_j G_j(R_1,\dots,R_n), $$ where $G_1(\mathbf u), \dots, G_n(\mathbf u) \in \mathbb K[[\mathbf u]]$ for $\mathbf u = (u_1,\dots,u_n)$.

Then, for any $F(\mathbf u) \in \mathbb K[[\mathbf u]]$ and $\mathbf a = (a_1,\dots,a_n)$ it holds that $$ [\mathbf x^\mathbf a] F(R_1,\dots,R_n) = [\mathbf u^\mathbf a] F(\mathbf u) G^\mathbf a(\mathbf u) \det\left(\delta_{ij} - \frac{u_j}{G_i(\mathbf u)} \frac{\partial G_i(\mathbf u)}{\partial u_j}\right), $$ where the expression under the determinant is the $(i,j)$-th element of the matrix.

For $G_i(\mathbf u) = \sum\limits_{j=1}^n a_{ij} u_j$ we get MMT.

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    $\begingroup$ A nice answer and a valuable enrichment to see how the problem can be solved. (+1) $\endgroup$ Dec 2, 2023 at 23:12
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I wish to add a more "from basics" proof here, to avoid Laurent series and so that this question can be closed as a duplicate.

This proof may be found at

Abdesselam, Abdelmalek, An algebraic independence result related to a conjecture of Dixmier on binary form invariants, Res. Math. Sci. 6, No. 3, Paper No. 26, 17 p. (2019). ZBL1442.13012.

I'll fill in the details.


Anyway, the idea is as follows. We first convert the given expression into something that is easier to work with. Then, we show that the expression, consisting of three variables, can be handled by reducing it to a combination of similar $2$-variable expressions, each of which can be computed with much more ease.

So, to get going, let's simplify the binomials. Note that all summations are over non-negative values of the summand, where we treat binomials as equal to $0$ if the lower index exceeds the upper index.

For every $k$, \begin{align*} &\binom{a+b}{a+k} \binom{b+c}{b+k}\binom{c+a}{c+k} \\ =& \frac{(a+b)!(b+c)!(c+a)!}{(a+k)!(a-k)!(b+k)!(b-k)!(c+k)!(c-k)!}\\ =& \frac{(a+b)!(b+c)!(c+a)!}{(2a)!(2b)!(2c)!} \frac{(2a)!}{(a-k)!(a+k)!} \frac{(2b)!}{(b-k)!(b+k)!} \frac{(2c)!}{(c-k)!(c+k)!} \\ = & \frac{(a+b)!(b+c)!(c+a)!}{(2a)!(2b)!(2c)!}\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k} \end{align*}

And therefore, $$ \sum_{k}(-1)^k \binom{a+b}{a+k} \binom{b+c}{b+k}\binom{c+a}{c+k} = \frac{(a+b)!(b+c)!(c+a)!}{(2a)!(2b)!(2c)!}\sum_{k} (-1)^k \binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}. $$

This is the desired rewrite. In the next section, we'll reduce the expression on the RHS to a two-variable version, which is easily computable.


We first recall the Chu-Vandermonde identity : for any $a,b,c>0$, $$\binom{a+b}{c} = \sum_{k} \binom{a}{k}\binom{b}{c-k}$$, where any binomial evaluates to $0$ if the lower index exceeds the upper index. This is proved by considering the number of subsets of size $c$ of $\{1,2,\ldots,a+b\}$. That's clearly the LHS, but the RHS asks for the number of subsets where exactly $k$ of them come from $\{1,\ldots,a\}$, summed over all non-negative $k$. Thus, both sides are counting the same set.

Now, we begin by considering the two variable expression and rewriting it : \begin{align*} &\binom{2a}{a+k}\binom{2b}{b+k}\\ =& \frac{(2a)!(2b)!}{(a+k)!(a-k)!(b+k)!(b-k)!} \\ =& \frac{(2a)!(2b)!}{(a+b)!(a-k)!(b+k)!}\binom{a+b}{b-k}. \\ = & \frac{(2a)!(2b)!}{(a+b)!(a-k)!(b+k)!}\binom{(a-k)+(b+k)}{b-k} \\ \overset{CV}{=} & \frac{(2a)!(2b)!}{(a+b)!(a-k)!(b+k)!}\sum_{l} \binom{a-k}{l}\binom{b+k}{b-k-l} \\ =& \sum_{l} \frac{(2a)!(2b)!(a-k)!(b+k)!}{(a+b)!(a-k)!(b+k)!l!(a-k-l)!(b-k-l)!(2k+l)!} \\ &= \frac{(2a)!(2b)!}{(a+b)!}\sum_{l} \frac{1}{l!(a-k-l)!(b-k-l)!(2k+l)!}. \end{align*}

Excellent. Now, we'll go back to the three variable variant and plug in the above rewrite immediately : \begin{align*} &\sum_{k} (-1)^k \binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k} \\ =& \frac{(2a)!(2b)!}{(a+b)!}\sum_{k}\sum_{l} \frac{(-1)^k}{l!(a-k-l)!(b-k-l)!(2k+l)!}\binom{2c}{c+k} \\ =& \frac{(2a)!(2b)!}{(a+b)!}\sum_{l}\sum_{k} \frac{(-1)^k}{l!(a-k-l)!(b-k-l)!(2k+l)!}\binom{2c}{c+k}, \end{align*} where the switch of summations is justified since the sum over $l$ is actually just a finite sum : $l \leq \min\{a,b\}$ must occur for the term to be non-zero, regardless of the value of $k$. Then, we get \begin{align*} &\frac{(2a)!(2b)!}{(a+b)!}\sum_{l}\sum_{k} \frac{(-1)^k}{l!(a-k-l)!(b-k-l)!(2k+l)!}\binom{2c}{c+k} \\ = & \frac{(2a)!(2b)!}{(a+b)!}\sum_{l}\sum_{k} \frac{(-1)^k}{(2k+2l)!(a-k-l)!(b-k-l)!}\binom{2k+2l}{l}\binom{2c}{c+k} \end{align*} Introduce a new variable $l' = l+k$. Then, we have \begin{align*} &\frac{(2a)!(2b)!}{(a+b)!}\sum_{l}\sum_{k} \frac{(-1)^k}{(2k+2l)!(a-k-l)!(b-k-l)!}\binom{2k+2l}{l}\binom{2c}{c+k} \\ =& \frac{(2a)!(2b)!}{(a+b)!}\sum_{l'}\sum_{k} \frac{(-1)^k}{(2l')!(a-l')!(b-l')!}\binom{2l'}{l'+k}\binom{2c}{c+k} \\ =& \frac{(2a)!(2b)!}{(a+b)!}\sum_{l'}\frac{1}{(2l')!(a-l')!(b-l')!}\sum_{k} (-1)^k\binom{2l'}{l'+k}\binom{2c}{c+k} \tag{*} \label{*} \end{align*}

We observe that the RHS is a two-variable version of the identity we want!! But is it easier to obtain? Indeed, it is, because we can use our two-variable binomial identity yet again. \begin{align*} &\sum_{k} (-1)^k\binom{2l'}{l'+k}\binom{2c}{c+k}\\ =& \sum_{k} (-1)^k \frac{(2l')!(2c)!}{(l'+c)!}\sum_{l} \frac{1}{l!(c-k-l)!(l'-k-l)!(2k+l)!}\\ =& \frac{(2l')!(2c)!}{(l'+c)!}\sum_l \frac{1}{(2k+2l)!(c-k-l)!(l'-k-l)!}\sum_{k} (-1)^k \binom{2k+2l}{l}. \end{align*} Again, let $l'' = k+l$ be the change of variable. Then, \begin{align*} &\frac{(2l')!(2c)!}{(l'+c)!}\sum_l \frac{1}{(2k+2l)!(c-k-l)!(l'-k-l)!}\sum_{k} (-1)^k \binom{2k+2l}{l}\\ &= \frac{(2l')!(2c)!}{(l'+c)!}\sum_{l''} \frac{1}{(2l'')!(c-l'')!(l'-l'')!}\sum_{k} (-1)^k \binom{2l''}{k}. \end{align*}

The sum over $k$ is well-known, though. Indeed, by a binomial expansion of $(1+(-1))^{2l''} = 0$, we see that $\sum_{k} (-1)^k \binom{2l''}{k} = 0$ for all $l'' > 0$, and $\sum_{k} (-1)^k \binom{2l''}{k} = 1$ if $l''=0$. Thus, in the above summation, only $l''=0$ survives, and we have $$ \frac{(2l')!(2c)!}{(l'+c)!}\sum_{l''} \frac{1}{(2l'')!(c-l'')!(l'-l'')!}\sum_{k} (-1)^k \binom{2l''}{k}\\ = \frac{(2l')!(2c)!}{(l'+c)!c!l'!}. $$

Now, we rush all the way back to \eqref{*} and see that \begin{align*} &\frac{(2a)!(2b)!}{(a+b)!}\sum_{l'}\frac{1}{(2l')!(a-l')!(b-l')!}\sum_{k} (-1)^k\binom{2l'}{l'+k}\binom{2c}{c+k} \\ =& \frac{(2a)!(2b)!}{(a+b)!}\sum_{l'}\frac{(2c)!}{(a-l')!(b-l')!(l'+c)!c!l'!}\\ =& \frac{(2a)!(2b)!(2c)!}{(a+b)!a!(b+c)!c!}\sum_l \binom{a}{l'}\binom{b+c}{b-l'}\\ \overset{CV}{=} & \frac{(2a)!(2b)!(2c)!}{(a+b)!a!(b+c)!c!}\binom{a+b+c}{b}\\ =& \frac{(2a)!(2b)!(2c)!(a+b+c)!}{(a+b)!(b+c)!(c+a)!a!b!c!}. \end{align*}

We have proved that $$ \sum_{k} (-1)^k \binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k} = \frac{(2a)!(2b)!(2c)!(a+b+c)!}{(a+b)!(b+c)!(c+a)!a!b!c!}. $$ Whence it follows from our first section calculations that $$ \bbox[yellow, border:2px solid red]{ \sum_{k} (-1)^k \binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k} = \frac{(a+b+c)!}{a!b!c!} } $$

This proof is elementary and involves nothing more than factorial manipulations and the Chu-Vandermonde identity. In fact, the proof itself was motivated by a request for a simple argument in this post which was solved by the poster itself. I only borrow the work and made some improvements to the flow which should make the argument easier to read.

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  • $\begingroup$ @SarveshRaichandranlyer: It seems when introducing $l^{\prime\prime}=k+l$ we have $\sum_{k} (-1)^k \binom{2l^{\prime\prime}}{l^{\prime\prime}-k}$. $\endgroup$ Dec 6, 2023 at 20:36

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