Closed form of sum. I tried to find a closed form for this :
$$\sum_{i = 1}^{d} \binom{d-i+n}{n-1}$$
I get this during my algebra task. Maybe there is combinatorial idea? 
 A: The given sum is the coefficient of $t^{n-1}$ in 
\begin{align*}
(1+t)^{n+d-1} & + (1+t)^{n+d-2} + \cdots + (1+t)^n \\
&= (1+t)^{n+d-1}\left(1+\frac{1}{1+t} + \cdots + \frac{1}{(1+t)^{d-1}}\right)\\
&= (1+t)^{n+d-1}\left(\frac{(1+t)^{d-1}+(1+t)^{d-2}+\cdots + 1}{(1+t)^{d-1}}\right)\\
&=(1+t)^n\left(\frac{1-(1+t)^d}{1-(1+t)}\right)\\
&= (1+t)^n \frac{(1+t)^d-1}{t} \\
&= \frac{(1+t)^{n+d}-(1+t)^n}{t} 
\end{align*}
and hence equals
$$\binom{n+d}{n} - 1$$
A: A counting proof:
Let us count the number of subsets with $n$ elements of $\{1,2,\ldots, n+d\}$. This is clearly $\binom{n+d}{n}$. 
We now count the subsets of size $n$ whose highest element is $n+d-i+1$, for $i = 1, 2, \ldots, d$. With $n+d-i+1$ already in the set, we need to choose additional $n-1$ elements from $\{1, 2, \ldots, n+d-i\}$ and hence this is $\binom{n+d-i}{n-1}$. This count misses the $n$ element subset $\{1,2,\ldots,n-1, n\}$ since at $i=d$, the highest element in the subset is $n+1$. Thus 
$$\sum_{i = 1}^{d} \binom{d-i+n}{n-1} = \binom{n+d}{n}-1$$
