Need to show following equality

Question

I want to show that the following equality holds for any $i,m, n \in \Bbb N$. I could not figure out how to show it analytically. Could you please help me? $$ \sum _{j=0}^n \binom{n}{j}\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}=\frac{1}{(i+1)\binom{m+1}{i+1}}$$

Answer 1

Use the sum of product of binomials formula

$\sum _{j=0}^k \begin{pmatrix}k-j \\r \end{pmatrix}\begin{pmatrix}i+j \\s \end{pmatrix}=\begin{pmatrix}k+i+1 \\r+s+1 \end{pmatrix}$

Let $k=m+n-i$, $r=m-i$, and $s=i$

$\sum _{j=0}^{n} \begin{pmatrix}m+n-i-j \\m-i \end{pmatrix}\begin{pmatrix}i+j \\i \end{pmatrix}=\begin{pmatrix}m+n+1 \\m+1 \end{pmatrix}$

where the index $j$ runs to $n$ since for $j>n$ the first binomial coefficient in the summation vanishes. Now it is just a matter of writing the binomial coefficients in terms of factorials and recombining

$$\sum _{j=0}^n \frac{(m+n-i-j)!}{(m-i)!(n-j)!}\frac{(i+j)!}{i!j!}=\frac{(m+n+1)!}{(m+1)!n!}$$

$$\sum _{j=0}^n \frac{n!}{(n-j)!j!}\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}=\frac{i!(m-i)!}{(m+1)!}$$

$$\sum _{j=0}^n \begin{pmatrix}n \\j \end{pmatrix} \frac{(i+j)!(m+n-i-j)!}{(m+n+1)!}=\frac{i!(m-i)!}{(m+1)!}=\frac{1}{(i+1)\begin{pmatrix}m+1 \\i+1 \end{pmatrix}}$$

Answer 2

We know from the Hypergeometric Distribution and the Vandermonde Identity that $$\large\sum_{j=0}^{n}\frac{\binom nj\binom m{r-j}}{m+n\choose r}=1$$ which is independent of $r$.

Summing the above from $r=0$ to $m+n$ gives $$\large\sum_{r=0}^{m+n}\sum_{j=0}^{n}\frac{\binom nj\binom m{r-j}}{m+n\choose r}=m+n+1$$

Put $i=r-j$ and note that values of $i>m$ results in zero terms, hence take summation from $i=0$ to $m$ only: $$\begin{align} \large\sum_{i=0}^m\sum_{j=0}^{n}\frac{\binom nj\binom mi}{m+n\choose i+j}&=m+n+1\\\\ \large\sum_{j=0}^{n}\frac{\binom nj\binom mi}{m+n\choose i+j}&=\frac{m+n+1}{m+1}\\\\ \large\sum_{j=0}^{n}\frac{\binom nj}{(m+n+1){m+n\choose i+j}} &=\frac 1{(m+1){\large\binom mi}}\\\\ \large\sum_{j=0}^{n}\binom nj\frac{(i+j)!(m+n-i-j)!}{(m+n+1)!} &=\frac 1{(i+1){\large\binom {m+1}{i+1}}}\qquad\blacksquare\\ \end{align}$$