A school has scheduled 3 volleyball games, 2 soccer games, and 4 basketball games. If you have a ticket that allows you to attend exactly 3 games and you decide to watch 2 basketball games and 1 game from the other sports, in how many different ways can you choose the games?

Introduction / Context:
This question mixes simple combinatorics with a real life scenario about choosing sports events. It tests your ability to correctly interpret phrases like "2 basketball games and 1 of the other events" and then apply combinations to count choices from different categories. Understanding how to handle selections from separate groups is important for many exam problems in which items are classified by type.

Given Data / Assumptions:

• Number of volleyball games: 3.
• Number of soccer games: 2.
• Number of basketball games: 4.
• You must attend exactly 3 games in total.
• Your choice must include 2 basketball games and exactly 1 non-basketball game (volleyball or soccer).
• Each game is distinct.

Concept / Approach:
We select games from two separate groups:

• First, choose 2 basketball games out of 4 available.
• Second, choose 1 game from the 3 volleyball and 2 soccer games, that is, from 5 non-basketball games.

In both cases, the order in which you choose games does not matter. Therefore, combinations are used for each selection, and then the counts are multiplied using the fundamental principle of counting.

Step-by-Step Solution:
Step 1: Number of ways to choose 2 basketball games from 4 is 4C2. Step 2: Compute 4C2 = 4! / (2! * 2!) = 6. Step 3: Total non-basketball games = 3 volleyball + 2 soccer = 5. Step 4: Number of ways to choose 1 non-basketball game from 5 is 5C1 = 5. Step 5: Use multiplication principle: total ways = 4C2 * 5C1 = 6 * 5. Step 6: Compute 6 * 5 = 30. Step 7: Therefore, there are 30 different ways to attend exactly 2 basketball games and 1 other game.

Verification / Alternative check:
You can list combinations more systematically in a smaller toy example to verify the structure and then trust the formula. For instance, if there were 3 basketball games and 2 others, you could manually list selections to see that the pattern 3C2 * 2C1 works correctly. Scaling back to the real numbers (4 and 5) keeps the logic identical, so the multiplication of separate combination counts is justified.

Why Other Options Are Wrong:
25: This might come from incorrectly doing 5C2, ignoring the condition about exactly 2 basketball games. 50: Could arise from mistakenly adding or multiplying extra factors without proper reasoning. 75: Too large and does not correspond to a valid combination structure for the given conditions. Only 30 matches the correct breakdown 4C2 * 5C1.

Common Pitfalls:
A common mistake is to combine all games and choose any 3 without enforcing the condition of exactly 2 basketball games. Another error is to treat volleyball and soccer separately and double count mixed selections. Always separate the problem into independent choices (here, basketball choice and non-basketball choice) and then use combinations for each selection. Finally, students sometimes confuse 5C1 with 5C2 or misapply permutations where the sequence of attending games is treated as important, which it is not in this question.

Final Answer:
The number of different ways to watch exactly two basketball games and one other game is 30.