Combinations and Permutations: Forming Committees with Specific Criteria

In this article, we'll explore the combinatorial mathematics behind forming a committee with specific gender criteria, and delve into the difference between permutations and combinations. The example we'll work through is forming a committee of 3 boys and 4 girls from a group of 6 boys and 5 girls.

Understanding the Problem

To form a committee of 3 boys from 6 boys, we need to understand that the order in which we choose the boys does not matter. This means we use combinations, not permutations. Similarly, for the 4 girls from 5 girls, the order is also irrelevant, so again, we use combinations.

Calculation Using Combinations

The formula for combinations is given by:

nCr n!/r!(n-r)!

Choosing 3 Boys from 6

We need to calculate the number of ways to choose 3 boys from 6. Using the combination formula:

$, binom{6}{3} frac{6!}{3!(6-3)!} frac{6 times 5 times 4}{3 times 2 times 1} 20.$

Choosing 4 Girls from 5

Similarly, we need to calculate the number of ways to choose 4 girls from 5:

$, binom{5}{4} frac{5!}{4!(5-4)!} frac{5}{1} 5.$

The total number of possible committees is the product of the number of ways to choose the boys and the number of ways to choose the girls:

$, text{Total Committees} binom{6}{3} times binom{5}{4} 20 times 5 100.$

Thus, the number of ways to form the committee is 100.

Alternative Approach: Permutations

We can also approach this problem using permutations. For the boys, we would calculate:

$, binom{6}{3} frac{6!}{3!(6-3)!} 120 / 6 20.$

For the girls, we would calculate:

$, binom{5}{4} frac{5!}{4!(5-4)!} 5 / 1 5.$

But since each girl chosen is distinct and order does not matter, we still arrive at the same 100 possible committees.

Additional Scenarios

Let's consider additional scenarios where the criteria change. If one boy rejects the offer, we have only 5 boys willing to serve. We now need to choose 3 boys from 5 and 4 girls from 5:

$, binom{5}{3} times binom{5}{4} 10 times 5 50.$

If we need at least 2 boys on the committee, we can have either 2 or 3 boys. We calculate the combinations for both scenarios:

2 Boys on the Committee

$, binom{6}{2} times binom{5}{2} 15 times 10 150.$

3 Boys on the Committee

$, binom{6}{3} times binom{5}{3} 20 times 10 200.$

The total number of arrangements where there are at least 2 boys on the committee is:

$, 150 200 350.$

Conclusion

Understanding the difference between permutations and combinations is crucial in solving problems related to committee formation. By applying the correct formula and method, we can find the total number of possible ways to form the committee, ensuring that the criteria are met.