# 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!

## 1 Answer

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.