chaotic_iak wrote: Suppose that $k,m,n$ are positive integers with $k \le n$. Prove that: \[\sum_{r=0}^m \dfrac{k \binom{m}{r} \binom{n}{k}}{(r+k) \binom{m+n}{r+k}} = 1\] The equation is equivalent to ${\sum_{r=0}^m \binom{r+k-1}{k-1} \binom{m+n-r-k}{n-k}}=\binom{m+n}{n}$ LHS is equal to the coefficient of $x^{k-1}y^{n-k}$ in polynomial $h(x,y)=\sum_{r=0}^{m+n-r-1}(1+x)^{r}(1+y)^{m+n-r-1}$. $h(x,y)=\frac{(1+x)^{m+n}-(1+y)^{m+n}}{x-y}=\sum_{i=1}^{m+n}\binom{m+n}i\frac{x^i-y^i}{x-y}$. The term $\frac{x^i-y^i}{x-y}$ contains $x^{k-1}y^{n-k}$ if and only if $i=n$. Now we see the coefficient of $x^{k-1}y^{n-k}$ in $h(x,y)$ is $\binom{m+n}{n}$. Q.E.D.

Using the facts $k \binom{n}{k} = n \binom{n-1}{k-1}$ and $(r+k) \binom{m+n}{r+k} = (m+n) \binom{m+n-1}{r+k-1}$, it can easily be obtained that it suffices to prove: $$\frac{n}{m+n} \sum_{r = 0}^{m} \frac{\binom{m}{r} \binom{n-1}{k-1}}{\binom{m+n-1}{r+k-1}} = 1.$$ After writing out all the factorials and simplifying, we just need: $$\sum_{r = 0}^{m} \binom{r+k-1}{k-1} \binom{m+n-k-r}{n-k} = \binom{m+n}{n}.$$ This can be interpreted combinatorially. Let's consider $m+n$ students lined up from left to right in a straight line. Label them $1, 2, \cdots, m+n$ from left to right. $n$ of them will be selected to form a committee. We will prove the identity above by casework on the position of the $k$th student from the left among those selected to the committee. Indeed, the label of the $k$th student must be $r+k$ for some $0 \le r \le m.$ The identity easily follows from here. $\square$

There are $m$ boys and $n$ girls in a classroom. We chose some of them one by one and stopped once we had $k$ girls. Therefore, we have chosen at least $k$ and at most $m+k$ students. In other words, the sum of the probabilities that there are $r$ boys and $k$ girls with the last one being a girl in a random sequence of $r+k$ students over $r$ is $1$. Hence the identity.

Hi could someone give some motivation of solving the by finding the coefficient of polynomial like in post #2 like by giving handout or tips thx

notice after we rewrite the condition we come to proving this $\sum^m_{r=0}\binom{r+k-1}{k-1}\binom{m+n-k-r}{n-k}=\binom{m+n}{n}$ notice the $\binom{m+n-k-r}{n-k}$ could be treated as extracting the coefficient of $x^{n-k}$ from polynomial $(1+x)^{m+n-k-r}$ $\sum^m_{r=0}\binom{r+k-1}{k-1}\binom{m+n-k-r}{n-k} =\sum^m_{r=0}[x^{n-k}](1+x)^{m+n-k-r}\binom{r+k-1}{r}=[x^{n-k}](1+x)^{m+n-k}\sum^{\infty}_{r=0}(-\frac{1}{1+x})^r\binom{-k}{r}=\sum^{\infty}_{r=0}(-\frac{1}{1+x})^r[u^r](1+u)^{-k}$ By subtitution rule it equal to $[x^{n-k}](1+x)^{m+n-k}*(1-\frac{1}{1+x})^{-k}$ so after simplifying we have $[x^n](1+x)^{m+n}$ and we are done Subtitution Rule $A(u) = \sum^{\infty}_{j=0}a_j*u^j = \sum^{\infty}_{j=0}u^j[z^j]A(z)$