Forming a Committee with Specific Gender Requirements

The problem of forming a committee of 6 members from a group of 7 boys and 4 girls with specific gender requirements involves using combinatorial mathematics. This article will outline the steps to solve the problem of forming a committee with at least 2 girls, and different interpretations of the problem when it involves both minimum and maximum female members.

Introduction to Committee Formation

Combinatorics is a fundamental branch of mathematics that deals with the selection, arrangement, and operation of elements in a set. In the context of forming a committee, combinatorial methods are used to determine the number of ways a committee can be formed under certain constraints. Specifically, we need to form a committee with at least 2 girls from a group of 7 boys and 4 girls.

Forming a Committee with at Least 2 Girls

To find the total number of ways to form a committee with at least 2 girls, we need to consider different sub-cases.

Cases to Consider

CASE 1: 2 Girls and 4 Boys CASE 2: 3 Girls and 3 Boys CASE 3: 4 Girls and 2 Boys

Calculating Each Case

Let's calculate the number of ways for each case one by one.

CASE 1: 2 Girls and 4 Boys

The number of ways to choose 2 girls from 4 is given by the binomial coefficient:

$$binom{4}{2} frac{4!}{2!2!} frac{4 cdot 3}{2 cdot 1} 6$$

The number of ways to choose 4 boys from 7 is:

$$binom{7}{4} frac{7!}{4!3!} frac{7 cdot 6 cdot 5}{3 cdot 2 cdot 1} 35$$

The total number of ways for Case 1 is:

$$6 times 35 210$$

CASE 2: 3 Girls and 3 Boys

The number of ways to choose 3 girls from 4 is:

$$binom{4}{3} 4$$

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

$$binom{7}{3} frac{7!}{3!4!} frac{7 cdot 6 cdot 5}{3 cdot 2 cdot 1} 35$$

The total number of ways for Case 2 is:

$$4 times 35 140$$

CASE 3: 4 Girls and 2 Boys

The number of ways to choose 4 girls from 4 is:

$$binom{4}{4} 1$$

The number of ways to choose 2 boys from 7 is:

$$binom{7}{2} frac{7!}{2!5!} frac{7 cdot 6}{2 cdot 1} 21$$

The total number of ways for Case 3 is:

$$1 times 21 21$$

Final Calculation

The total number of ways to form a committee with at least 2 girls is the sum of the totals from all cases:

$$210 140 21 371$$

Therefore, the total number of ways to form a committee of 6 members with at least 2 girls is 371.

Interpreting the Problem Different Ways

The problem can be interpreted in different ways, requiring a more detailed analysis to derive the correct solution.

Exact 2 Girls with No Additional Conditions

If the committee must have exactly 2 girls and can have any combination of boys, the number of ways to select the girls is:

$$binom{4}{2} 6$$

The number of ways to select the boys is:

$$binom{7}{4} 35$$

The total number of ways to form the committee is:

$$6 times 35 90$$

At Least Two Girls and At Most Two Girls

If the committee must have at least 2 girls and at most 2 girls, then we need to consider the following scenarios:

At Least 2 Girls

This gives us:

2 girls and 4 boys (from above, it is 90 ways) 3 girls and 3 boys (as calculated earlier, it is 140 ways) 4 girls and 2 boys (15 ways)

The total number of ways for at least 2 girls is:

$$90 140 15 245$$

At Most 2 Girls

This gives us:

2 girls and 4 boys (90 ways) 1 girl and 5 boys (4 ways from girls and 6 from boys, which is 4 times 6 24 ways) 0 girls and 6 boys (1 way)

The total number of ways for at most 2 girls is:

$$90 24 1 115$$

To summarize, the committee can be formed in 245 ways if it has at least 2 girls and at most 2 girls.