From
a group of 12 people, how many ways can you form a committee of 5 people such
that the committee contains at least 2 women and at least 1 person with prior
experience in a similar role?
Solve it with using permutation and combination

To solve the problem of forming a committee of 5 people from a group of 12, with the constraints that the committee must contain at least 2 women and at least 1 person with prior experience in a similar role, we can approach it using combinations.

### Step 1: Define the Groups

1. **Total people (12 people):**
   - Suppose there are \( W \) women and \( M = 12 - W \) men.
   - Let \( E \) be the number of people with prior experience, and \( N = 12 - E \) be those without prior experience.

### Step 2: Use Inclusion-Exclusion Principle

The problem requires us to count the number of ways to form a committee of 5 people such that it contains at least 2 women and at least 1 person with prior experience.

#### Total combinations without any restrictions:
The total number of ways to select 5 people out of 12 is given by:
\[
\binom{12}{5} = \frac{12!}{5!(12-5)!} = 792
\]

#### Exclude cases where conditions are not met:

1. **Case 1: Committees with fewer than 2 women**
   - **0 women**: Select all 5 members from men (or \( 12-W \) group):
     \[
     \binom{12-W}{5}
     \]
   - **1 woman**: Select 1 woman and 4 men:
     \[
     \binom{W}{1} \times \binom{12-W}{4}
     \]

2. **Case 2: Committees with no experienced members**:
   - Select all 5 members from those without prior experience:
     \[
     \binom{12-E}{5}
     \]

### Step 3: Calculate the Desired Combinations

Using the inclusion-exclusion principle, the number of valid committees is given by:
\[
\text{Valid committees} = \binom{12}{5} - \text{(invalid due to fewer women)} - \text{(invalid due to no experience)}
\]

We don't have exact numbers for \( W \) and \( E \), so we’ll outline the steps you would take to compute:

- Calculate the combinations for \( \binom{12-W}{5} \) and \( \binom{W}{1} \times \binom{12-W}{4} \).
- Calculate \( \binom{12-E}{5} \).
- Subtract these invalid combinations from the total \( \binom{12}{5} = 792 \).

### Example Calculation:
If \( W = 6 \) and \( E = 4 \):

1. Invalid due to fewer women:
   - \( 0 \) women: \( \binom{6}{5} = 0 \) (as you cannot choose 5 men from 6).
   - \( 1 \) woman: \( \binom{6}{1} \times \binom{6}{4} = 6 \times 15 = 90 \).

2. Invalid due to no experience:
   - \( \binom{8}{5} = 56 \) (if \( 4 \) people have experience, 8 don't).

Then:
\[
\text{Valid committees} = 792 - 90 - 56 = 646
\]

### Final Answer
The number of ways to form a committee that meets the given criteria is 646.

Would you like further details on any step, or do you have any questions? 

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Here are 5 relative questions:
1. How would the calculation change if the number of people with prior experience increases?
2. What happens if we allow committees with fewer members, say 3 or 4?
3. How do the calculations change if we add more restrictions, like limiting the number of men?
4. How does the inclusion-exclusion principle apply in combinatorial problems in general?
5. Can this problem be extended to handle larger groups or more complex constraints?

**Tip:** When dealing with combinatorial problems, breaking them down into cases (like we did here) is often helpful in managing complex constraints.

Learn how to solve combinatorial problems involving forming committees with specific criteria. This example focuses on forming a committee of 5 people from a group of 12, ensuring at least 2 women and at least 1 person with prior experience. Detailed step-by-step solution using combinations and the inclusion-exclusion principle.

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