Combinatorics using Linear Transformation > Determine the number of integer solutions to $x_1+x_2+x_3+x_4 \le 72$ such that: > > * $1 \le x_1 \le 12$ > * $0 \le x_2\le 10$ > * $3 \le x_3\le 13$ > * $5 \le x_4\le 36$ > At first, I introduced a variable $x_5$ such that $x_1+x_2+x_3+x_4+x_5=72$ and applied a linear transformation such that: * $y_1 = x_1-1$ * $y_2 = x_2$ * $y_3 = x_3-3$ * $y_4 = x_4-5$ * $y_5 = x_5$ and then regarranging and substituting values into the original equation, I get $y_1+y_2+y_3+y_4+y_5=63$ such that: * $y_1 \le 12$ * $y_2 \le 10$ * $y_3 \le 13$ * $y_4 \le 36$ * $y_5 \ge 0$ my instincts tell me I should apply another linear transformation, but I'm not quite sure how to go from here. Any tips would be appreciated!

Any choice of $(x_1,x_2,x_3,x_4)$ that satify the constraints \begin{eqnarray*} 1 \le x_1 \le 12 \\\ 0 \le x_2\le 10 \\\ 3 \le x_3\le 13 \\\ 5 \le x_4\le 36 \end{eqnarray*} will satify the constraint $x_1+x_2+x_3+x_4 \le 72$ so there are $12$ choices for $x_1$,$11$ choices for $x_2$,$11$ choices for $x_3$,$32$ choices for $x_4$, so there are $12 \times 11 \times 11 \times 32 = \color{red}{46464}$ solutions.