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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.

 May 14, 2022
 #1
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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 

 May 14, 2022
 #3
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Hey, if you try and list all the possible additions --- you might miss some... 

Guest May 22, 2022
 #2
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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 :)

 May 22, 2022
edited by Guest  May 22, 2022

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