1995 MathCounts State Team #8 In professional football, it is possible to score 2 points (for a safety), 3 points (for a field goal), or 6 points (for a touchdown). If a touchdown is scored, it is possible to score 1 additional point, The score of a football game is 21 to 17. Find the total number of possible combinations of points to achieve this final score.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2222096#p2222096
How do you solve the problem? Can you give me a solution that's not so confusing? (The linked solution uses the floor function, and I don't see the relevance of the floor function.)
 A: 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.
A: In professional football it is possible to score 2 points (for a safety), 3 points (for a field goal), or 6 points (for a touchdown). If a touchdown is scored, it is possible to score one additional point. The score of a football game is 21 to 17. Find the total number of possible combinations to achieve this final score.
Your answer is correct.  I, too, would use organized casework...
Note that we have two odd scores available $(7$ and $3).$  And we have two even scores available $(6$ and $2).$
We must use an odd number of odd numbers to get a total that is odd.  Let's list all such totals that are less than or equal to 21.  To make sure I didn't miss any, I started with the largest number of 7's, and worked my way down.
So, the total number of ways of achieving a football score of $21$ to $17$ is $18\times 10=\boxed{180}.$
