How many solutions to the $a+b+c+d=100$ exist? Given a,b,c,d belonging to the set of whole numbers and given the equation
$a+b+c+d=100$
How many solutions like
$(a,b,c,d)=(80,10,5,5)\; ; \; (a,b,c,d)=(0,1,2,97)$
exist? We can repeat elements and the order does not matter.

We're looking for the non-negative integer solutions for the equation $x_1 + x_2 + x_3 + x_4 = 100$.
Instead of using numbers for writing the solutions, we will use strokes, so for instance we represent the solution $ x_1 = 1, x_2 = 1, x_3 = 1, x_4 =97 $, or 1 + 1 + 1 + 97, like this: | + | + | + ||| $ \cdots$ [97 strokes].
Now, each possible solution is an arrangement of 100 strokes and 3 plus signs, so the number of arrangements is P(103; 100, 3) =
$\frac{103!}{100! *3!}$
The general solution for such questions -combinations with repitition- is:
$\mathit{P}(n+r-1;r,n-1) = \frac{(n+r-1)!}{r!(n-1)!}=\binom{n+r-1}{r}$
