One thing that will matter quite a bit is whether the order matters. Well, let's see if we can poke around a bit at the answer. Suppose we wanted to find out how many combinations there were of 3 whole numbers that add to 6 or less. I think we could list them:
0+0+0
0+0+1
0+0+2
0+0+3
0+0+4
0+0+5
0+0+6
0+1+0 (Here you can see, by comparing with the second line, that I'm assuming order matters.)
0+1+1
0+1+2
0+1+3
0+1+4
0+1+5
0+2+0
0+2+1
0+2+2
0+2+3
0+2+4
0+3+0
0+3+1
0+3+2
0+3+3
0+4+0
0+4+1
0+4+2
0+5+0
0+5+1
0+6+0
1+0+0
At this point, we are starting to see a pattern. Up until the last line, I'd say we had 7+6+5+4+3+2+1 = 28 ways to write it. If we change the first digit to a 1, then each possibility is going to go down by 1. So we'd have 6+5+4+3+2+1 = 21 ways to write it. These are the Triangular Numbers, denoted by $T_n$. The definition of the Triangular Number $T_n$ is that
$$T_n=1+2+3+\dots+n,$$
and thanks to Gauss, we know that
$$T_n=\frac{n(n+1)}{2}.$$
To get our final result, we must sum the Triangular Numbers from $T_7$ down to $T_1$. That is, the answer to this smaller question is
$$N=\sum_{j=1}^7 T_j=\sum_{j=1}^7 \frac{j(j+1)}{2}=84.$$
In general,
$$N=\sum_{j=1}^n T_j=\frac16 n(n+1)(n+2).$$
This is the number of ways to get a sum of $n$ or less from three whole numbers where order matters.
But now, supposing we were to change the number of numbers to 4. How would that change our result? Well, for each value $f$ of the fourth number, we'd have $N-f$ ways to write the new sum. So, we'd have to do
$$\sum_{j=1}^7 \left[\frac16 n(n+1)(n+2)\right].$$
So, how would this work for 16 numbers?
Moreover, if we need to eliminate the idea that order matters, how could we correct these (too large) values?
