Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $

How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$

I tried the following:

We use the falling factorial power: $$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$ so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$

Then

$${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} = \frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$

And

$$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$ $$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$ $$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$ $$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$

According to the Binomial-coefficients:

$$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$ $$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$

What is wrong ? und How can I continue? :/

Solution 1:

Using Negative Binomial Coefficients and Vandermonde's Identity, we get $$ \begin{align} \sum_{k=0}^n\binom{x+n-k-1}{n-k}\binom{y+k-1}{k} &=\sum_{k=0}^n(-1)^{n-k}\binom{-x}{n-k}(-1)^k\binom{-y}{k}\tag{1}\\ &=(-1)^n\sum_{k=0}^n\binom{-x}{n-k}\binom{-y}{k}\tag{2}\\ &=(-1)^n\binom{-x-y}{n}\tag{3}\\ &=\binom{n+x+y-1}{n}\tag{4} \end{align} $$ Explanation:
$(1)$: Negative Binomial Coefficient conversion
$(2)$: algebra
$(3)$: Vandermonde's Identity
$(4)$: Negative Binomial Coefficient conversion

Solution 2:

Hint: The binomial formula with the Cauchy product \begin{align*} (x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k} \end{align*} does not use falling factorials $x^{\underline{k}}$ resp. $y^{\underline{n-k}}$.

Here is a step by step answer similar to that by @MarkoRiedel. It's convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} \binom{n}{k}=[z^k](1+z)^n \end{align*}

We obtain

\begin{align*} \sum_{k=0}^{n}&\binom{x-1+n-k}{n-k}\binom{y-1+k}{k}\\ &=\sum_{k=0}^{\infty}\binom{x-1+n-k}{n-k}\binom{-y}{k}(-1)^k\tag{1}\\ &=\sum_{k=0}^\infty [t^{n-k}](1+t)^{x-1+n-k}[z^k](1+z)^{-y}(-1)^k\tag{2}\\ &=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty(-1)^kt^k(1+t)^{-k}[z^k](1+z)^{-y}\tag{3}\\ &=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty\left(-\frac{t}{1+t}\right)^k[z^k](1+z)^{-y}\\ &=[t^n](1+t)^{x-1+n}\left(1-\frac{t}{1+t}\right)^{-y}\tag{4}\\ &=[t^n](1+t)^{x+y-1+n}\tag{5}\\ &=\binom{x+y-1+n}{n} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial identity $\binom{-p}{q}(-1)^q=\binom{p+q-1}{q}$ and we extend the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

• In (2) we apply the coefficient of operator twice.

• In (3) we do some rearrangements by using the linearity of the coefficient of operator and we also use the rule \begin{align*} [z^{p-q}]A(z)=[z^p]z^{q}A(z) \end{align*}

• In (4) we apply the substitution rule \begin{align*} A(t)=\sum_{k=0}^\infty a_kt^k=\sum_{k=0}^\infty t^k[z^k]A(z)\\ \end{align*} with $z=-\frac{t}{1+t}$.

• In (5) we do some simplifications.

• In (6) we select the coefficient from $t^n$.

Solution 3:

We have $$\sum_{k=0}^n {y-1+k\choose k} {x-1+n-k\choose n-k} = \sum_{k=0}^n [z^{k}] \frac{1}{(1-z)^y} [z^{n-k}] \frac{1}{(1-z)^x} \\ = [z^n] \frac{1}{(1-z)^y} \frac{1}{(1-z)^x} = [z^n] \frac{1}{(1-z)^{x+y}} ={x+y-1+n\choose n}.$$