<s>[INST] How many integer solutions to the equation (discrete math) How many integer solutions to the equation $x_1 + x_2 + x_3 = 15$ so that $x_1 \ge 3, x_2 \ge 2$ and $x_3 \ge 0$ ? I honestly don't know where to start on this
Thanks
 [/INST] Let's consider the number of solutions of $x_1 + x_2 + x_3 = n$, where $x_1 \in \{3, 4, ...\}$, $x_2 \in \{2, 3, ...\}$ and $x_3 \in \mathbb{N} \cup \{0\}$.
Let $a_n$ be the number of solutions of that equation. You're looking for $a_{15}$.
We can easily find the generating function of $a_n$:
$$g(x) = (x^3 + x^4 + ...)(x^2 + x^3 + ...)(1+x+x^2+...)$$
So that the coefficient of $x^n$ is $a_n$:
$$g(x) = \displaystyle{\sum_{n=0}^\infty a_n x^n}$$
Therefore, for $|x|<1$, we conclude that:
$$g(x) = \dfrac{x^3}{1-x} \cdot \dfrac{x^2}{1-x} \cdot \dfrac{1}{1-x} = x^5 \cdot (1-x)^{-3}$$
Using the Generalized Binomial Theorem, and since $\binom{-\alpha}{k} = \binom{\alpha + k - 1}{k}(-1)^k$:
$$g(x) = x^5 \displaystyle{\sum_{k=0}^\infty \binom{-3}{k} (-1)^k x^k} = x^5 \displaystyle{\sum_{k=0}^\infty \binom{2+k}{k}x^5} = \displaystyle{\sum_{k=0}^\infty \dfrac{(k+1)(k+2)}{2}x^{k+5}}$$
Therefore:
$$
a_n = \left \{
      \begin{matrix} 
         0 & n < 5 \\
         \dfrac{(k+1)(k+2)}{2} & n \geq 5      \end{matrix}
   \right .
$$
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