How many integer solutions?

Principle of inclusion and exclusion is used in this proof.

Let $\begin{cases}x_1 &=& y_1+3 \\ x_2 &=& y_2 \\ x_3 &=& y_3+7 \\ \end{cases}$ .

The expression becomes $y_1+y_2+y_3=18$ with $0\le y_1\le 6, \, 0\le y_2\le 8, \, 0\le y_3 \le 10$ .

The non-negative integer solutions to the equation $y_1+y_2+y_3=18$ is ${20 \choose 2}=190$ .

Let $a_1=7, a_2=9$ and $a_3=11$ . Let $T_k=\{(y_1, y_2, y_3): y_1+y_2+y_3=18, y_i\text{ are non-negative integers and }y_k\ge a_k \}$ for $k=1,2,3$ .

Now $\left|T_1 \right|={13 \choose 2}$ , $\left|T_2 \right|={11 \choose 2}$ , $\left|T_3 \right|={9 \choose 2}$ , $\left|T_1 \cap T_2 \right|={4 \choose 2}$ , $\left|T_1 \cap T_3 \right|={2 \choose 2}$ and $\left|T_2 \cap T_3 \right|= \left|T_1 \cap T_2 \cap T_3 \right|=0$ .

By Principle of inclusion and exclusion, $\left|T_1 \cup T_2 \cup T_3 \right| = \left|T_1 \right|+\left|T_2 \right|+\left|T_3 \right|-\left|T_1 \cap T_2 \right|-\left|T_1 \cap T_3 \right|-\left|T_2 \cap T_3 \right|+ \left|T_1 \cap T_2 \cap T_3 \right|=162$ .

Hence the number of integer solutions to the equation (with the conditions) is $190-162=\boxed{28}$ .