A nice identity involving urns and balls problem Prove the identity: $$\frac{\displaystyle\sum_{k=0}^{a} {n+a-k-2\choose n-2}}{\displaystyle {n+a-1\choose a}} = 1$$ where $C_{i}^{j}$ is defined as the number of ways to simultaneously choose $j$ objects from $i$ objects.
My attempt: I was trying to use a combinatorial argument by saying that out of the given $n$ balls, each of the terms within the summation is the probability of a given urn containing exactly $k$ balls for $k = 0,1,2,...,n$. Thus, the sum reflects the sum of all the probabilities of that urn having exactly $k$ balls, so it must be $1$. 
My question: I would like to see an algebraic proof for this identity. Could someone please help with such a proof? In case my argument above is incorrect, please help point out the mistake.
 A: Here is a variation based upon the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
  [x^k](1+x)^n=\binom{n}{k}
  \end{align*}

We obtain
  \begin{align*}
\sum_{k=0}^n\binom{n+a-k-2}{n-2}&=\sum_{k=0}^n\binom{-n+1}{a-k}(-1)^{a-k}\tag{1}\\
&=\sum_{k=0}^\infty[x^{a-k}](1+x)^{-n+1}(-1)^{a-k}\tag{2}\\
&=(-1)^a[x^a](1+x)^{-n+1}\sum_{k=0}^\infty (-x)^k\tag{3}\\
&=(-1)^a[x^a](1+x)^{-n}\tag{4}\\
&=(-1)^a\binom{-n}{a}\tag{5}\\
&=\binom{n+a-1}{a}\tag{6}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$ with $p=n-1$ and $q=a-k$.

*In (2) we apply the coefficient of operator.

*In (3) we use the linearity of the coefficient of operator and apply the rule
$$[x^{p-q}]A(x)=[x^p]x^qA(x)$$ We also extend the upper range of the series to $\infty$ without changing anything since we are adding zeros only.

*In (4) we use the geometric series expansion.

*In (5) we select the coefficient of $x^a$.

*In (6) we use again the binomial identity as we did in (1).
A: ${n+a-1 \choose n-1}={n+a-1 \choose a}$ is the number of non-negative integer solutions to:
$x_1+x_2+\ldots+x_n=a$
Let $x_1=k$ with $k \in \{0,1,\ldots,a\}$. Then the corresponding number of solutions is ${n+a-k-2 \choose n-2}$.
