How many solutions for the equation $t_1+t_2+t_3+t_4=6$

Question

How many solutions are there to the equation $t_1+t_2+t_3+t_4=6$ if $t_i\in \{-1,0,1,2,3,4,5,6,7,8,9\}$ for all $i=1,2,3,4$?

I know how to answer this problem without the set of numbers $t_i$. But when I should keep the set $t_i$ in mind, I don't really know how to solve it. I know the answer might be smaller but again, I have no clue.

My idea is to use the formula: $C(n+r-1,r)$ and if I ignore the set, then there would be $C(4+6-1,6)=C(9,6)=84$ ways but I have to use the set, any help?

Answer 1

Problem can be rephrased as:

How many solutions are there for $s_1+s_2+s_3+s_4=10$ if $s_i\in\{0,1,2,3,4,5,6,7,8,9,10\}$?

Here we applied $s_i=t_i+1$.

Fortunately the demands that $s_i\geq0$ for $i=1,2,3,4$ together with $s_1+s_2+s_3+s_4=10$ allready force the $s_i$ not to exceed $10$, so another rephrase is possible:

How many solution are there for $s_1+s_2+s_3+s_4=10$ if the $s_i$ are nonnegative integers?

To solve that you can use stars and bars, leading to $$\binom{10+4-1}{4-1}$$ possibilities.

Answer 2

You can define $u_i=t_i+1$ and now you are looking for the ways to solve $u_1+u_2+u_3+u_4=10$ with all the $u_i$ nonnegative. The maximum is not a problem here as no solution has a $u_i \gt 10$ so use your usual stars and bars.