This problem is just like giving 13 cookies to 5 kids. We need to say how many of the 13 units go to each of the 5 variables. In other words, we have 13 stars and 4 bars (the bars are like the “+” signs in the equation).
If \(x_i\) can be 0 or greater, we are in the standard case with no restrictions. So 13 stars and 4 bars can be arranged in \({17 \choose 4}\) ways.
Now each variable must be at least 1. So give one unit to each variable to satisfy that restriction. Now there are 8 stars left, and still 4 bars, so the number of solutions is \({12 \choose 4}\text{.}\)
Now each variable must be 2 or greater. So before any counting, give each variable 2 units. We now have 3 remaining stars and 4 bars, so there are \({7 \choose 4}\) solutions.