How do you solve $x_1+x_2+x_3+x_4 \le 45$, where all the $x$'s are nonnegative with the added conditions that $x_1 \ge 0, x_2\ge 1, x_3 \ge 4,$ and $x_4 \le 5$? I know how to do the normal stars and bars argument, but the inequality and these conditions make it more complicated.
Combinatorics: stars and bars with conditions
combinatorics
Best Answer
By using slack variables. Find the number of solutions of $x_1+x_2+x_3+x_4+x_5=45$, where each $x_i$ is a non-negative integer. That converts the last inequality into an equality.
For $x_2$ and $x_3$, just use the constraints as your starting point: $x_1+y_2+y_3+x_4+x_5 = 40$, where $y_2+1=x_2$ and $y_3+4=x_3.$
To handle the inequality on $x_4$, use the technique discussed above to ascertain the number of solutions of your original equation where $x_4 \geq 6: x_1+y_2+y_3+y_4+x_5=34,$ where $y_4+6=x_4$. Subtract this number from the number of solutions of your first equation.
Thus, your answer should be $\binom{44}{4}-\binom{38}{4}.$