Question
How many different committees of 5 reporters can be formed with at least 2 women from the given city?
Answers
GPT-4o mini
To solve the problem of how many different committees of 5 reporters can be formed with at least 2 women, we need to consider various combinations of men and women in these committees.
Assuming we have a certain number of men (M) and women (W) reporters, the possible compositions of committees of 5 reporters with at least 2 women could be:
1. 2 Women and 3 Men
2. 3 Women and 2 Men
3. 4 Women and 1 Man
4. 5 Women
We will calculate the number of committees for each case separately and then sum them up.
Let’s denote:
- \( C(n, r) \) as the binomial coefficient representing the number of ways to choose \( r \) items from \( n \) items.
For the calculations, we will need the actual numbers of men and women. However, since these numbers are not provided, I will express the solutions in terms of \( M \) and \( W \).
### Case 1: 2 Women and 3 Men
The number of ways to choose 2 women from \( W \) and 3 men from \( M \) is given by:
\[
C(W, 2) \cdot C(M, 3)
\]
### Case 2: 3 Women and 2 Men
The number of ways to choose 3 women from \( W \) and 2 men from \( M \):
\[
C(W, 3) \cdot C(M, 2)
\]
### Case 3: 4 Women and 1 Man
The number of ways to choose 4 women from \( W \) and 1 man from \( M \):
\[
C(W, 4) \cdot C(M, 1)
\]
### Case 4: 5 Women
The number of ways to choose 5 women from \( W \) (assuming \( W \geq 5 \)):
\[
C(W, 5)
\]
### Total Committees
The total number of committees with at least 2 women is the sum of all the cases:
\[
\text{Total Committees} = C(W, 2) \cdot C(M, 3) + C(W, 3) \cdot C(M, 2) + C(W, 4) \cdot C(M, 1) + C(W, 5)
\]
### Conclusion
You will need to substitute the actual numbers of men (\( M \)) and women (\( W \)) reporters into the formula to obtain the total number of different committees.
Assuming we have a certain number of men (M) and women (W) reporters, the possible compositions of committees of 5 reporters with at least 2 women could be:
1. 2 Women and 3 Men
2. 3 Women and 2 Men
3. 4 Women and 1 Man
4. 5 Women
We will calculate the number of committees for each case separately and then sum them up.
Let’s denote:
- \( C(n, r) \) as the binomial coefficient representing the number of ways to choose \( r \) items from \( n \) items.
For the calculations, we will need the actual numbers of men and women. However, since these numbers are not provided, I will express the solutions in terms of \( M \) and \( W \).
### Case 1: 2 Women and 3 Men
The number of ways to choose 2 women from \( W \) and 3 men from \( M \) is given by:
\[
C(W, 2) \cdot C(M, 3)
\]
### Case 2: 3 Women and 2 Men
The number of ways to choose 3 women from \( W \) and 2 men from \( M \):
\[
C(W, 3) \cdot C(M, 2)
\]
### Case 3: 4 Women and 1 Man
The number of ways to choose 4 women from \( W \) and 1 man from \( M \):
\[
C(W, 4) \cdot C(M, 1)
\]
### Case 4: 5 Women
The number of ways to choose 5 women from \( W \) (assuming \( W \geq 5 \)):
\[
C(W, 5)
\]
### Total Committees
The total number of committees with at least 2 women is the sum of all the cases:
\[
\text{Total Committees} = C(W, 2) \cdot C(M, 3) + C(W, 3) \cdot C(M, 2) + C(W, 4) \cdot C(M, 1) + C(W, 5)
\]
### Conclusion
You will need to substitute the actual numbers of men (\( M \)) and women (\( W \)) reporters into the formula to obtain the total number of different committees.