Beggar method or something else

We can solve equation X+Y+Z+W= 23

So no of solution = 26C3

Let's define each subproblem $SP(n)$ as follows:

"Find the number of non-negative solutions of the equation $X + Y + Z = n$ ."

Solving one subproblem with the aid of stars and bars method : $SP(n) = {n \ + \ 2 \choose 2 }$ .

Now, solving the main problem, the answer is: $\sum\limits_{0}^{23} SP(i) = \sum\limits_{0}^{23} {i \ + \ 2 \choose 2 } = {2 \choose 2 } + {3 \choose 2 } + \ ... \ + {24 \choose 2 } + {25 \choose 2 }$ .

Actually, using the Hockey-Stick Identity in Pascal's Triangle, we can make it easier to calculate, which gives us: $\sum\limits_{0}^{23} {i \ + \ 2 \choose 2 } = {26 \choose 3 }$ .

So finally, the answer is ${26 \choose 3 } = 2600$ .