Kim,Megan, Nina,and Paula are planning to go to a movie theater, where three different movies-A,B,and C—are being shown simultaneously in different auditoriums. Each of them will see only one of the movies but not necessarily the same movie as any of the other three people.One possible outcome is that Kim and Megan will see C while Nina and Paula will see A. How many possible outcomes are there?

To solve this problem, we need to determine how many possible ways the four people (Kim, Megan, Nina, and Paula) can be assigned to one of the three movies (A, B, or C).

### Step 1: Understanding the Problem

Each person can independently choose one of the three movies (A, B, or C). Therefore, for each person, there are 3 options.

### Step 2: Counting the Total Possible Outcomes

Since each of the 4 people has 3 choices, the total number of outcomes can be calculated by multiplying the number of choices for each person:

\[
3 \times 3 \times 3 \times 3 = 3^4 = 81
\]

### Step 3: Conclusion

The total number of possible outcomes is 81.

Would you like more details, or do you have any questions?

### Related Questions

1. How would the answer change if only two movies were available instead of three?
2. What if one person decides not to watch any movie?
3. How many outcomes are there if at least two people must watch the same movie?
4. How would the solution change if all four had to watch the same movie?
5. What if there are five friends choosing between three movies?
6. How many outcomes are possible if each person can choose between three movies or stay home?
7. How many outcomes are there if only two people can watch the same movie?
8. What is the probability that all four friends choose the same movie?

### Tip:
For counting problems involving independent choices, the total number of outcomes is often a product of the number of options available for each choice.

Learn how to count the number of possible outcomes when Kim, Megan, Nina, and Paula choose between three different movies shown simultaneously in a theater. This problem involves combinatorial principles and independent choices. Suitable for students in Grades 7-9.

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