Question

How many ways can a committee be selected consisting of two Independents, two Republicans, and two Democrats if the choices are made from seven Independents, nine Republicans, and eight Democrats?

Short answer

21,168 ways.

Step-by-step solution

Understanding the problem

We need to form a committee that includes 2 Independents, 2 Republicans, and 2 Democrats. The possible selections are from a pool of 7 Independents, 9 Republicans, and 8 Democrats. The task is to calculate the number of different ways to form this committee.

Calculate the combinations for Independents

To choose 2 Independents from 7, use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Here, \(n = 7\) and \(r = 2\). \[ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \] There are 21 ways to choose 2 Independents from 7.

Calculate the combinations for Republicans

To choose 2 Republicans from 9: \[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \] There are 36 ways to choose 2 Republicans from 9.

Calculate the combinations for Democrats

To choose 2 Democrats from 8: \[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \] There are 28 ways to choose 2 Democrats from 8.

Combine the combinations

The number of ways to form the complete committee is the product of the individual choices (since each choice is independent). Multiply the combinations: \[ 21 \times 36 \times 28 = 21,168 \] There are 21,168 ways to form the committee according to the given criteria.

Key concepts

Committee Selection

Selecting a committee is a classic application of combinatorial counting, where you need to form a group with specific characteristics from a larger pool of individuals. In this exercise, you are asked to form a committee of six members, with specific quotas for Independents, Republicans, and Democrats. Here are some key points to consider when selecting a committee:

• The total number of spots available in the committee.
• The requirements or constraints for each category, such as the number of members needed from each group.
• The total number of candidates available in each category to choose from.

To solve this exercise, you need to break down the problem by category and compute how many ways you can select members from each category. After calculating these separate possibilities, you multiply them together to find the total number of ways to form the desired committee. This approach ensures you are considering all possible combinations according to the set rules.

Combinations

Combinations refer to the selection of items from a larger set without regard for the order of selection. This is different from permutations, where order does matter. In this exercise, when selecting 2 Independents from 7, 2 Republicans from 9, and 2 Democrats from 8, the order doesn't matter, and thus combinations is the correct approach.The formula for combinations is denoted as \( \binom{n}{r} \), which stands for "n choose r." It calculates the number of ways to select \( r \) items from a set of \( n \) items. The formula is:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]

• \( n! \) ("n factorial") means multiplying all whole numbers from \( n \) down to 1.
• The divide by \( r!(n-r)! \) adjusts for the fact that order is irrelevant.

Employing combinations helps you count distinct groups without repeating sequences, and it's ideal for tasks like committee selections where roles are equal.

Binomial Coefficient

The binomial coefficient \( \binom{n}{r} \) is a key mathematical concept used in problems involving combinations. It is used to determine the number of ways to choose \( r \) items from \( n \) items without regard to the order of selection. The binomial coefficient is inherently tied to the calculation of combinations and central to many combinatorial problems.Here’s how it works:

• When you see \( \binom{7}{2} \), it means "7 choose 2," representing the different ways you can select 2 items from a total of 7.
• The calculation involves factorials, such as \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \).
• You divide \( 7! \) similarly after adjusting for the specific "choose" number \( r \).

Binomial coefficients simplify various combinatorial counting challenges, enabling quick computation of selections. Using such coefficients not only provides the answer but a deeper understanding of how many combinations are possible in seemingly complex selection problems.