Number of solutions discrete math

Question

I asked this on the main site and was told to ask here. How many solutions does the equation $x_1 +x_2 +x_3 = 8$ have with integers $x_i \ge 0$ with the additional restrictions: $x_1 \ge 2$ and $x_3 \ge 3$? Also with different restrictions of $x_1 \le 2$ and $x_3 \le 3$. I was able to figure it out with the restrictions but can’t figure out how to do it with them. I got $\binom{10}{7}$ I believe for without restrictions. I know I have to use the binomial coefficient again just don’t know the values to use.

Answer 1

You can set $y_1 = x_1 + 2$ and $y_3 = x_3 + 3$ and then you need to find how many solutions there are for $y_1 + x_2 + y_3 = 3$, where all the summands are non-negative. By the Stars and Bars principle you have $\binom{5}{2}$ such combinations. Hence there are $\binom{5}{2} = 10$ ways to write $x_1+x_2+x_3 =8$ with $x_1 \ge 2$ and $x_3 \ge 3$.

For the second case things get a bit more complicated, as you need to use the Inclusion-Exclusion Prinicple. First there are $\binom{10}{2}=45$ solutions without any restriction. Then find out how many solutions are there s.t. $x_1 > 2$. By using the same argument as above you will get $\binom{7}{2} = 21$ such combinations. Similarly there are $\binom{6}{2} = 15$ combinations with $x_3 > 3$. Also there $\binom{3}{2} = 3$ combinations where both $x_1 > 2$ and $x_3 > 3$. Finally the wanted number is:

$$45-21-15+3 = 12$$