Summation of a factorial (total number terms in a polynomial) By induction I can prove : 
$$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!}  = \frac{(D+M)!}{D!M!} $$
However, I couldn't derive the right hand side directly.
It would be of great help if anyone can solve it!!
 A: $(1+x)^{M} (1+x)^{D} = (1+x)^{M+D} $
compute coefficient of $x^{D}$ on both sides. $$$$
on LHS, it is $$\sum^{M}_{t=0}\frac{(t+D-1)!}{t!(D-1)!} $$ 
and on it is RHS $$ \frac{(D+M)!}{D!M!} $$
A: Start with the binomial coefficient relation 
$${n\choose D}={n-1\choose D}+{n-1\choose D-1}$$ 
rewritten as 
$${n\choose D}-{n-1\choose D}={n-1\choose D-1}.$$ 
Adding these increments gives
$$ {D+M\choose D}-{D-1\choose D}=\sum_{n=D}^{D+M}{n-1\choose D-1}.$$
Since ${D-1\choose D}=0$ this gives you the sum that you want.
This technique is called upper summation, see also here.

Here's another combinatorial argument to complement Norbert's. 
The number of ways to select $D$ distinct values from the set $\{1,2,\dots,D+M\}$ is $${D+M\choose D}.$$ For $0\leq t\leq M$, the number of such selections with maximum value $D+t$ is
$${D-1+t\choose D-1}.$$
A: Here is a combinatorial proof.
Assume you have $D+1$ types of cakes, and you allowed to choose $M$ cakes. Of course you can take several cakes of the same type. The amount of possible choices is combination with repetitions:
$$
{(D+1)+M - 1 \choose D}=\frac{(M+D)!}{M!D!}\tag{1}
$$
How can we classify all possible choices? We say our choice belongs to the $t$-th class if we choose $M-t$ cakes of the first type. How many choices of $t$-th type are there? Well this is amount of possible choices of remaining $t$ cakes from remaining $D$ types of cakes, which is equal to 
$$
{D+t-1 \choose D-1}=\frac{(D+t-1)!}{(D-1)!t!}
$$
Since we are allowed to take only $M$ cakes, there are $M+1$ classes - $t$ varies from $0$ to $M$. Now we see that total amount of choices is sum of amount of choices over all classes:
$$
\sum\limits_{t=0}^M{D+t-1\choose D-1}=\sum\limits_{t=0}^M\frac{(D+t-1)!}{(D-1)!t!}
$$
On the other hand this amount is equals to $(1)$.
