Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

A committee of 3 men and 2 women is to be chosen from a group of 12 men and 8 women. Determine the number of different ways of selecting the committee.

Step-by-Step Solution

Answer

There are 6,160 different ways to select the committee.

Step 1: Determine Number of Ways to Choose Men

We need to choose 3 men from a total of 12 men. The number of ways to choose 3 men from 12 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. Therefore, we calculate it as: \[ \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220. \]

Step 2: Determine Number of Ways to Choose Women

Next, we need to choose 2 women from a total of 8 women. Using the combination formula again, \( \binom{8}{2} \), we calculate: \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28. \]

Step 3: Calculate Total Number of Ways to Form the Committee

The number of different ways to form the committee is the product of the number of ways to choose the men and the number of ways to choose the women. Therefore, we multiply the results from Step 1 and Step 2: \[ 220 \times 28 = 6160. \]

Key Concepts

Using the Combination Formula

To determine the number of ways to form a committee, we rely on a fundamental concept in combinatorics known as the combination formula. The combination formula helps us calculate the number of ways to select a subset of elements from a larger set, where the order of selection does not matter.

Mathematically, the combination formula is represented as \( \binom{n}{r} \), where \( n \) is the number of items to choose from and \( r \) is the number of items you want to choose. This formula is derived from the expression:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Here, \( n! \) (n factorial) is the product of all positive integers up to \( n \). For instance, to choose 3 men from a pool of 12, we calculate \( \binom{12}{3} \), resulting in 220 possible combinations. Similarly, choosing 2 women from 8 involves calculating \( \binom{8}{2} \), which gives 28 combinations.

The combination formula is a cornerstone of combinatorial mathematics, as it systematically accounts for the multiplicity of selections needed in any non-sequential scenario.

Permutations and Combinations

It is essential to understand the distinction between permutations and combinations when tackling problems involving selection. Permutations concern arrangements where order matters, while combinations focus on selections where order does not matter.

Consider a simple example: choosing 2 out of 3 different fruits. In permutations, apple-orange is different from orange-apple; thus, the number of permutations is greater due to varied ordering. However, in combinations, apple-orange and orange-apple are seen as the same selection.

• Permutations: Concerned with order.
• Combinations: Order is irrelevant, primary focus of our problem.

In the committee selection problem, we’re using combinations to find different groups because we only care about who is on the committee, not the order they're listed. Understanding these differences is crucial to applying the proper mathematical approach, ensuring accuracy in solving problems involving group selection.

Simplifying Mathematical Calculations

When working on combinatorial problems, simplifying mathematical calculations can save time and reduce errors. This involves understanding factorials and how they simplify in the combination formula.

Start by recognizing that factorial \( n! \) grows rapidly with even small increases in \( n \). While computing these directly for large numbers can be daunting, canceling common terms often reduces complexity. For example, the expression \( \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \) efficiently simplifies factorial calculations, avoiding large multiplications.

• Focus on canceling terms in the numerator and the denominator.
• Utilize recognizable patterns seen in smaller numbers to rationalize expressions.
• Practice with smaller numbers to build confidence in manipulating expressions.

This strategic simplification is not only useful in the classroom but also in various professional fields requiring combinatorial analysis. By breaking down complex expressions into simpler components, students can efficiently tackle a wide range of mathematical problems.