In American football, a touchdown scores 7 points, a field goal scores 3 points, and a safety scores 2 points. If a team scored a total of 24 points, how many combinations of plays is possible? The order of the plays does not matter, so touchdown and field goal in either order is considered the same combination.

Guest May 14, 2022

#1**0 **

2 + 3 + 7 + 2 + 3 + 7

7 + 3 + 7 + 7

2 + 3 + 3 + 3 + 2 + 2 + 3 + 3 + 3

3 + 3 + 3 + 3 + 3 + 3 + 3 + 3

2 + 2 + 7 + 2 + 7 + 2 + 2

2 + 2 + 3 + 2 + 2 + 2 + 2 + 2 + 7

3 + 3 + 2 + 3 + 3 + 3 + 7

3 + 2 + 2 + 2 + 2 + 3 + 2 + 2 + 2 + 2 + 2

8 ways

Guest May 14, 2022

#2**0 **

We want to try to minimize the cases to minimize mistakes. We have 4 cases:

Case1 = no touchdowns

Case2 = 1 touchdown

Case3 = 2 touchdowns

Case4 = 3 touchdowns

Case4: If there are 3 touchdowns, then that means there is one possible way since you must have 3 touchdowns and one field goal.

Case4 = 1 way

Case3: If there are 2 touchdowns, we are left with 10 points. 3x + 2y = 10. If x = 0, then y = 5. If x = 1, then y is invalid. If x = 2, then y = 2. If x = 3, y is invalid. Thus, there are 2 ways if for Case3.

Case3 = 2 ways

Case2: If there is 1 touchdown, we are left with 17 points. 3x + 2y = 17. If x = 0, then y is invalid. If x = 1, then y = 7. If x = 2, then y is invalid. If x = 3, then y = 4. (We notice that x has to be odd) If x = 5, then y = 1. If x = 7, then y is negative --- impossible...

Case2 = 3 ways.

Case1: If there are no touchdowns, we have to make do with 24 points, 3x + 2y = 24. x = 0, y = 12. (x must be even) x = 2, y = 9. x = 4, y = 6. x = 6, y = 3. x = 8, y = 0. Case1 has 5 cases (x = 0, 2, 4, 6, 8).

Case1 = 5 ways.

Hence, we have 1 + 2 + 3 + 5 ways = **11 ways.**

**proyaop :)**

proyaop May 22, 2022

edited by
Guest
May 22, 2022