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Math  /  Algebra

Question3. At the end of yesterday's soccer game between Team Why and Team Zed, Team Why had scored 3 goals and Team Zed had scored 2 goals. At half-time of the game, Team Why had scored yy goals and Team Zed had scored zz goals. If y≥0y \geq 0 and z≥0z \geq 0, how many possibilities are there for the ordered pair of integers (y,z)(y, z) ? (In soccer, each team's score is always a non-negative integer that never decreases as the game proceeds.)

Studdy Solution

STEP 1

1. Team Why scored a total of 3 goals by the end of the game.
2. Team Zed scored a total of 2 goals by the end of the game.
3. At half-time, Team Why had scored y y goals and Team Zed had scored z z goals.
4. y y and z z are non-negative integers.
5. The scores do not decrease as the game proceeds.

STEP 2

1. Determine the possible values for y y .
2. Determine the possible values for z z .
3. Count the number of valid ordered pairs (y,z)(y, z).

STEP 3

Since Team Why scored 3 goals in total, and scores do not decrease, y y can be any integer from 0 to 3.
Possible values for y y are: 0,1,2,3 0, 1, 2, 3 .

STEP 4

Since Team Zed scored 2 goals in total, and scores do not decrease, z z can be any integer from 0 to 2.
Possible values for z z are: 0,1,2 0, 1, 2 .

STEP 5

Count the number of valid ordered pairs (y,z)(y, z) by combining the possible values of y y and z z .
For each value of y y , there are 3 possible values for z z .
Total possibilities = 4×3=12 4 \times 3 = 12 .
The number of possibilities for the ordered pair (y,z)(y, z) is:
12 \boxed{12}

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