I am sorry Alan, I just don't get it.
If I want to know how many ways I can put 3 identical b***s 9 different boxes such that there was a maximum of 1 ball in each box then I would say
1. 9C3=84 or
2. $$\frac{9*8*7}{3!}=84$$
Now , you have said

So $$1+3+6+10+15+21+28 =$$ $$\frac{9*8*7}{3!}=84$$
-----------------------------
OR

I am really struggling with this Alan.
I totally understand that the answer to this is 84 (doing it my way) but I just can't understand the logic that you use to do it your way.
ALSO can you show me how to expand this triple summation please. I have no idea :(
I have always had trouble with sigma notation. As soon as it is above the most basic level I get lost :/
Maybe you could throw me a couple of double summations that I could try on my own. :)