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.Find the number of formations that can form a team.  How many types of five people can have at most one man?

### Step 1: Calculate the Total Number of Combinations

First, let's find the total number of ways to select five students from a class of twelve. As previously mentioned, this can be calculated using the combination formula:

\[
\binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792
\]

So, the total number of ways to form a team of five students from twelve is **792**.

### Step 2: Calculate the Number of Teams with at Most One Man

Now, let's consider the scenario where the team has at most one man. This means the team can have either:
- No men (all women)
- Exactly one man

Assume the class has \(m = 6\) men and \(w = 6\) women. Let's calculate each case.

#### Case 1: No Men (All Women)

For the team to consist entirely of women, you select all five members from the six women:

\[
\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6 \times 5!}{5! \times 1} = 6
\]

So, there are 6 ways to form a team with no men.

#### Case 2: Exactly One Man

For a team with exactly one man, you need to choose 1 man from 6 and 4 women from 6:

\[
\text{Number of ways} = \binom{6}{1} \times \binom{6}{4}
\]

Calculate each part:

\[
\binom{6}{1} = 6 \quad \text{(ways to choose 1 man)}
\]

\[
\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \quad \text{(ways to choose 4 women)}
\]

So, the number of ways to form a team with exactly one man is:

\[
6 \times 15 = 90
\]

### Step 3: Add the Results

Finally, sum the cases:

\[
\text{Total ways with at most one man} = 6 \text{ (no men)} + 90 \text{ (one man)} = 96
\]

### Summary:
1. **Total number of teams:** 792
2. **Number of teams with at most one man:** 96

Would you like further clarification on this solution, or do you have other questions?

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Here are some related questions:
1. What is the number of ways to form a team with at least two men?
2. How would the solution change if the class had 8 women and 4 men?
3. How does the concept of complementary counting apply in this scenario?
4. What if the team needed to have exactly three men?
5. How does this problem illustrate the use of binomial coefficients in combinatorics?

**Tip:** When solving combinatorial problems, breaking the problem into smaller cases (like "at most one man") can simplify the calculation and ensure accuracy.

Explore how to calculate the number of ways to form a team of five from twelve students, including scenarios with at most one man. Learn the application of combinatorics and binomial coefficients in solving this problem.

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