Counting Committees with at Least One Female Member

In many organizational structures, it's common to form committees consisting of a mix of individuals, often with particular demographics requirements. This article explores how to select a committee of 5 people from 7 boys and 6 girls, ensuring that the committee includes at least one girl. Utilizing the principle of complementary counting, we'll derive a straightforward formula for achieving this requirement.

Introduction

The principle of complementary counting is a useful approach in combinatorics, which involves counting the number of valid outcomes by first calculating the total number of possible outcomes and then subtracting the number of invalid outcomes. In our scenario, we're tasked with forming a committee of 5 people from a pool of 7 boys and 6 girls, with the constraint that each committee must have at least one girl.

Principle of Complementary Counting

To apply the principle of complementary counting, we first calculate the total number of ways to select 5 people from the 13 individuals (7 boys and 6 girls). This can be done using the binomial coefficient:

$$ binom{13}{5} frac{13!}{5!(13-5)!} frac{13!}{5! cdot 8!} 1287 $$

Next, we calculate the number of ways to select a committee of 5 people that consists of only boys. Since there are 7 boys, this can be calculated as:

$$ binom{7}{5} frac{7!}{5!(7-5)!} frac{7!}{5! cdot 2!} 21 $$

Using complementary counting, we subtract the number of all-boy committees from the total number of possible committees:

$$ text{Number of committees with at least one girl} binom{13}{5} - binom{7}{5} 1287 - 21 1266 $$

Key Distinctions and Combinations

We can further break down the possible compositions of the committee into distinct categories:

{6C2 7C3} : 2 women and 3 men {6C3 7C2} : 3 women and 2 men {6C4 7C1} : 4 women and 1 man {6C5 7C0} : 5 women and 0 men

The calculation for each category is as follows:

2 Women and 3 Men

$$ binom{6}{2} times binom{7}{3} 15 times 35 525 $$

3 Women and 2 Men

$$ binom{6}{3} times binom{7}{2} 20 times 21 420 $$

4 Women and 1 Man

$$ binom{6}{4} times binom{7}{1} 15 times 7 105 $$

5 Women and No Men

$$ binom{6}{5} times binom{7}{0} 6 times 1 6 $$

Summing these values gives us the total number of valid committees:

$$ 525 420 105 6 1056 $$

Conclusion

In summary, the total number of ways to select a committee of 5 people from 7 boys and 6 girls, with at least one girl, is 1056. This approach effectively leverages the principle of complementary counting, ensuring a more manageable breakdown of the problem into simpler calculations.

Related Keywords

Combinatorics, committee selection, principle of complementary counting, binomial coefficient, permutations, combinations.