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Let's solve the first problem clearly.

### Problem 1: 
A committee of 6 is to be chosen from 10 men and 7 women such that it contains at least 3 men and 2 women. However, two particular women refuse to serve on the same committee. How many different ways can this be done?

#### Step-by-step solution:

**Step 1: Understanding the conditions**

- The committee consists of **6 people**.
- The committee must contain at least **3 men** and **2 women**.
- There are **10 men** and **7 women** available.
- Two particular women **cannot** serve on the same committee.

**Step 2: Identify possible gender combinations for the committee**

Since the committee must contain at least 3 men and 2 women, the valid combinations of men and women are:

- **3 men and 3 women**.
- **4 men and 2 women**.
- **5 men and 1 woman**.

**Step 3: Calculate the number of ways to choose men and women for each combination**

1. **For 3 men and 3 women**:
   - Number of ways to choose 3 men from 10 men = \( \binom{10}{3} \).
   - Number of ways to choose 3 women from 7 women = \( \binom{7}{3} \).
   
   However, two women refuse to serve together. Let's denote them as \(W_1\) and \(W_2\). We need to subtract the cases where both \(W_1\) and \(W_2\) are chosen together.
   
   - Number of ways to choose 3 women without restriction = \( \binom{7}{3} \).
   - Number of ways to choose 3 women where both \(W_1\) and \(W_2\) are included: We must select 1 more woman from the remaining 5 (since \(W_1\) and \(W_2\) are already chosen), which is \( \binom{5}{1} \).
   
   Hence, the total number of valid ways to choose 3 women is:
   \[
   \binom{7}{3} - \binom{5}{1}
   \]
   
2. **For 4 men and 2 women**:
   - Number of ways to choose 4 men from 10 men = \( \binom{10}{4} \).
   - Number of ways to choose 2 women from 7 women = \( \binom{7}{2} \).
   
   We need to exclude the cases where \(W_1\) and \(W_2\) are both chosen.
   - Number of ways to choose 2 women without restriction = \( \binom{7}{2} \).
   - Number of ways to choose 2 women where \(W_1\) and \(W_2\) are both chosen = 1 (since they are the only 2 women).
   
   Hence, the total number of valid ways to choose 2 women is:
   \[
   \binom{7}{2} - 1
   \]

3. **For 5 men and 1 woman**:
   - Number of ways to choose 5 men from 10 men = \( \binom{10}{5} \).
   - Number of ways to choose 1 woman from 7 women = \( \binom{7}{1} \).
   
   In this case, there are no restrictions on the woman chosen because only 1 woman is selected, so we don't have to worry about \(W_1\) and \(W_2\) both being chosen.

**Step 4: Calculate the total number of ways for each scenario**

1. **3 men and 3 women**:
   \[
   \binom{10}{3} \times \left( \binom{7}{3} - \binom{5}{1} \right)
   \]

2. **4 men and 2 women**:
   \[
   \binom{10}{4} \times \left( \binom{7}{2} - 1 \right)
   \]

3. **5 men and 1 woman**:
   \[
   \binom{10}{5} \times \binom{7}{1}
   \]

**Step 5: Add the results from all the scenarios**

The total number of ways to form the committee is the sum of the valid choices from all three cases.

If you'd like, I can calculate the numerical values for you or walk you through each combination calculation. Would you like more details?

A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee?

Learn how to solve a combinatorics problem involving the selection of a committee of 6 people, containing at least 3 men and 2 women, with specific constraints. Detailed step-by-step explanation using combination formulas and set theory. Suitable for undergraduate discrete mathematics students.

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