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 y0y \geq 0 and z0z \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
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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