Forming Committees with Restrictions: A Comprehensive Guide

Introduction:

In this guide, we will explore how to solve combinatorial problems, specifically the formation of committees with varying restrictions. We will delve into the process of forming a committee of 2 women and 3 men from a group of 5 women and 7 men, and then introduce a restriction that complicates the problem. By the end of this guide, you will understand the mathematical principles behind these calculations and be able to apply them to similar scenarios.

Forming Committees Without Restrictions

Let's start with the basic problem of forming a committee consisting of 2 women and 3 men from a group of 5 women and 7 men. We will use the concept of combinations to solve this.

Choosing Women:

The number of ways to choose 2 women from 5 can be calculated using the combination formula (binom{n}{r}), which is defined as:

(binom{n}{r} frac{n!}{r!(n - r)!})

For choosing 2 women from 5:

(binom{5}{2} frac{5!}{2!(5 - 2)!} frac{5 times 4}{2 times 1} 10)

Choosing Men:

The number of ways to choose 3 men from 7 is:

(binom{7}{3} frac{7!}{3!(7 - 3)!} frac{7 times 6 times 5}{3 times 2 times 1} 35)

Total Committees:

The total number of committees can be calculated by multiplying the number of ways to choose the women and the men:

(text{Total Committees} binom{5}{2} times binom{7}{3} 10 times 35 350)

Addressing the Restriction: Feuding Men

Now, let’s introduce a restriction: two of the men, M1 and M2, are feuding and refuse to serve on the committee together. We need to calculate the number of valid committees under this new condition.

Committees Including Both M1 and M2:

If M1 and M2 are both on the committee, we need to choose 1 more man from the remaining 5 men:

(binom{5}{1} 5)

Thus, the number of ways to choose 2 women and include both M1 and M2 is:

(text{Committees with M1 and M2} binom{5}{2} times binom{5}{1} 10 times 5 50)

Committees Not Including Both M1 and M2:

To find the number of valid committees, we subtract the number of committees that include both M1 and M2 from the total number of committees calculated earlier:

(text{Valid Committees} text{Total Committees} - text{Committees with M1 and M2} 350 - 50 300)

Final Answer:

Thus, the number of different committees consisting of 2 women and 3 men where the two feuding men do not serve together is 300.

Conclusion

This guide has provided a step-by-step solution to forming a committee with restrictions, using combinatorial mathematics. By understanding the principles of combinations and how to apply restrictions, you can solve similar problems efficiently.

Keywords:

Combination, Committee Formation, Combinatorial Mathematics