Originally Posted by
Boysilver
Yes, I didn't explain very clearly. In your example, what I'm saying is that I'd like to prove that the number of ways in which we can find non-negative integers $\displaystyle a_0,a_1,a_2,a_3,a_4,a_5$ (also, previously I said "positive integers" - I meant non-negative integers) such that
$\displaystyle a_0+a_1+a_2+a_3+a_4+a_5<3$
is 28 possibilities.
(I said "$\displaystyle \ldots < n$" last time - I meant $\displaystyle <t$. Sorry again!)
Just for clarity, I'll list them here:
$\displaystyle (a_0,a_1,a_2,a_3,a_4,a_5) =$ (0,0,0,0,0,0), (1,0,0,0,0,0), (1,0,0,0,0,1), (1,0,0,0,1,0), (1,0,0,1,0,0), (1,0,1,0,0,0), (1,1,0,0,0,0), (0,1,0,0,0,0), (0,1,0,0,0,1), (0,1,0,0,1,0), (0,1,0,1,0,0), (0,1,1,0,0,0), (0,0,1,0,0,0), (0,0,1,0,0,1), (0,0,1,0,1,0), (0,0,1,1,0,0), (0,0,0,1,0,0), (0,0,0,1,0,1), (0,0,0,1,1,0), (0,0,0,0,1,0), (0,0,0,0,1,1), (0,0,0,0,0,1), (2,0,0,0,0,0), (0,2,0,0,0,0), (0,0,2,0,0,0), (0,0,0,2,0,0), (0,0,0,0,2,0), (0,0,0,0,0,2).
So there are indeed 28. I'd like a way of proving that the number of ways will be $\displaystyle t+n \choose t-1$ in the general case for any t and n.