Before the question is closed, I would like to give you a start.
Suppose $a+b+c+d=N$. Let us look at $\binom{N}{a}\binom{N-a}{b}\binom{N-a-b}{c}\binom{N-a-b-c}{d}$ (the last term is $1$, it is just there to make things look nice.)
Calculate, using the usual formula for $\binom{n}{k}$.
The first term is $\frac{N!}{a!(N-a)!}.$
The second term is
$\frac{(N-a)!}{b!(N-a-b)!}.$
The third term is
$\frac{(N-a-b)!}{c!(N-a-b-c)!}.$
Note that $N-a-b-c=d$.
Multiply, and observe the very nice cancellations! We get
$\frac{N!}{a!b!c!d!}.$
The "general" case solution is basically the same, except that all those subscripts tend to make things less obvious.
Added: By "the usual formula" for $\binom{n}{k}$ I mean
$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$
Since the question has not yet been closed, I am adding a link to a Wikipedia entry which I think is quite well written, and which I hope you will give you all of the additional information you may need.