Counting Committees with Specific Gender Distribution
Introduction
In combinatorics, the problem of forming a committee from a specific pool of candidates is a common task. This article specifically tackles the problem of determining the number of ways a committee of 5 people can be chosen from a group of 6 men and 3 women, ensuring that the committee composition is balanced in a specific way.
Selection of Committee Members
We wish to determine the number of ways to form a committee of 5 people from a group consisting of 6 men and 3 women, with the requirement that the committee includes a specific number of women and men. This leads us to consider different scenarios based on the number of women and men in the committee.
Combinatorial Analysis
The committee can have 2 women and 3 men, 3 women and 2 men, 4 women and 1 man, or 5 women and 0 men. We will analyze each of these scenarios separately using combinatorial methods.
Case 1: 2 Women and 3 Men
The number of ways to select 2 women from 3 women is given by the binomial coefficient ( C(3, 2) ). Similarly, the number of ways to select 3 men from 6 men is ( C(6, 3) ).
Mathematically, this can be expressed as:
[ text{Number of ways} C(3, 2) times C(6, 3) ]
Substituting the values:
[ text{Number of ways} frac{3!}{2!(3-2)!} times frac{6!}{3!(6-3)!} 3 times 20 60 ]
Case 2: 3 Women and 2 Men
The number of ways to select 3 women from 3 women is given by the binomial coefficient ( C(3, 3) ), and the number of ways to select 2 men from 6 men is ( C(6, 2) ).
Mathematically, this can be expressed as:
[ text{Number of ways} C(3, 3) times C(6, 2) ]
Substituting the values:
[ text{Number of ways} frac{3!}{3!(3-3)!} times frac{6!}{2!(6-2)!} 1 times 15 15 ]
Case 3: 4 Women and 1 Man
The number of ways to select 4 women from 3 women is given by the binomial coefficient ( C(3, 4) ), which is 0 since it's not possible to select 4 items from 3. Similarly, the number of ways to select 1 man from 6 men is ( C(6, 1) ).
Mathematically, this can be expressed as:
[ text{Number of ways} C(3, 4) times C(6, 1) ]
Substituting the values:
[ text{Number of ways} 0 times 6 0 ]
Case 4: 5 Women and 0 Men
The number of ways to select 5 women from 3 women is given by the binomial coefficient ( C(3, 5) ), which is 0 since it's not possible to select 5 items from 3. Similarly, the number of ways to select 0 men from 6 men is ( C(6, 0) ).
Mathematically, this can be expressed as:
[ text{Number of ways} C(3, 5) times C(6, 0) ]
Substituting the values:
[ text{Number of ways} 0 times 1 0 ]
Total Number of Committees
Summing up all the possible ways from the different cases, we get:
[ text{Total number of committees} 60 15 0 0 75 ]
Conclusion
This analysis shows that there are 75 different ways to form a committee of 5 people from a group of 6 women and 7 men, ensuring that the committee includes a balanced number of women and men in the specific distribution required.
The key takeaway from this analysis is that each selection method needs to be considered independently, and the combinatorial methods using binomial coefficients provide a clear and concise way to count the total number of possible committees.