I'll help you enumerate the ways you can score $17$ points, and let you work out how many ways there are to score $21$ points (or any other amount you are interested in).
If you score $17$ points, you can have scored at most $2$ touchdowns (since $3$ touchdowns would yield at least $18$ points). Let's consider the various cases:
Case A (2 touchdowns scored)
Here, we have to consider the possibility that $0$, $1$, or $2$ of the extra points after the touchdown were converted. If $2$ extra points were converted, that yields $14$ points. Thus, you must kick a field goal to get the remaining $3$ points. Similarly, with $1$ extra point you must convert $2$ safeties and with $0$ extra points, you must convert $1$ safety and $1$ field goal. Using TD, EP, FG, and saf as abbreviations, the three ways to score $17$ points with $2$ touchdowns are:
- $2$ TDs, $2$ EPs, $1$ FG;
- $2$ TDs, $1$ EP, $2$ saf;
- $2$ TDs, $1$ FG, $1$ saf.
Case B (1 touchdown scored)
First, consider the case where you have converted the extra point after the touchdown. This accounts for $7$ points, so you have $10$ remaining points to be distributed among field goals and safeties. If you have $a$ field goals and $b$ safeties, you will score $3a + 2b$ points, so you are trying to solve $3a + 2b = 10$ with $a$ and $b$ both non-negative integers. Since $3a = 10 - 2b$ is an even number, $a$ must be an even number. But $a$ cannot be bigger than $10/3$, so only $a = 0$ (in which case $b = 5$) and $a = 2$ (in which case $b = 2$) work. This gives two more possibilities:
- $1$ TD, $1$ EP, $5$ saf;
- $1$ TD, $1$ EP, $2$ FG, $2$ saf.
You also need to consider if the extra point was not scored after the touchdown. I'll leave that as an exercise for you. The method is similar to the analysis in the previous paragraph.
Case C ($0$ touchdowns scored)
Now you are trying to solve $3a + 2b = 17$, and you can deduce that $a$ must be odd. Only $a = 1$, $3$, and $5$ work, giving you three more possibilities. Again, I'll leave the rest of the analysis for you.
