A class of 12 students is assigned to select five students from twelve to work as a team on a special project.  Find the number of ways that a team of five members can be formed.Case 2 :   Team without both people.
 Let's say two members of a group don't get along and refuse to work together as a team.  Find the number of formats in which teams can be organized according to these conditions?

To solve this problem, we need to calculate the number of ways to select a team of 5 students from a class of 12, considering two different cases:

### Case 1: Total Number of Teams
First, let's calculate the total number of ways to form a team of 5 members from 12 students.

This is given by the combination formula:
\[
\text{Total number of teams} = \binom{12}{5}
\]
Where \(\binom{12}{5}\) is the number of combinations of 12 students taken 5 at a time.

### Calculation:
\[
\binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792
\]
So, the total number of possible teams of 5 members is \( 792 \).

### Case 2: Teams without Both People
Now, let’s find the number of ways to form a team of 5 members such that the two specific students (say \( A \) and \( B \)) who don't get along are not both included in the team.

#### Step 1: Teams with Both \( A \) and \( B \)
First, calculate the number of teams that include both \( A \) and \( B \).
- If both \( A \) and \( B \) are included, we need to select 3 more students from the remaining 10 students.
\[
\text{Number of teams with both \( A \) and \( B \)} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120
\]

#### Step 2: Teams without Both \( A \) and \( B \)
To find the number of teams without both \( A \) and \( B \), subtract the number of teams that include both from the total number of teams:
\[
\text{Number of teams without both \( A \) and \( B \)} = \binom{12}{5} - \binom{10}{3} = 792 - 120 = 672
\]

### Final Answer:
The number of ways to form a team of 5 members without both \( A \) and \( B \) is \( \boxed{672} \).

Would you like more details or have any questions?

Here are some related questions to explore:
1. What if the team size changes to 6 members?
2. How does the solution change if three members refuse to work together?
3. How would you calculate the number of teams with exactly one of \( A \) or \( B \)?
4. What if \( A \) and \( B \) must be in the team together instead?
5. How does the problem change if there are more than 12 students?

**Tip:** Always consider different cases when members of a group have specific restrictions or preferences; this approach often simplifies counting problems.

Learn how to calculate the number of ways to form a team of 5 members from a group of 12 students, considering different conditions such as excluding specific members. This problem involves combinatorics and binomial coefficients. Suitable for students in Grades 10-12.

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