Problem 67 - A Graphical Approach to College Algebra

Question

Use any or all of the methods described in this section to solve each problem. Committee Choices In a club with 8 men and 11 women members, how many 5 -member committees can be chosen that have the following? (a) All men (b) All women (c) 3 men and 2 women (d) No more than 3 women

Step-by-Step Solution

Answer

(a) 56, (b) 462, (c) 3080, (d) 8526

Step 1: Understanding the Problem

We need to find the number of ways to form different types of committees from a club which includes 8 men and 11 women. We will use combinations, as the order in which committee members are chosen does not matter.

Step 1: Calculate Committees with All Men

For a committee of 5 all men to be formed from 8 men, we select 5 out of 8. The combination formula is used here: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Thus, the number of ways to choose 5 men from 8 is: \[\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.\]

Step 2: Calculate Committees with All Women

For a committee of 5 all women to be formed from 11 women, we select 5 out of 11. Thus, the number of ways to choose 5 women from 11 is: \[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462.\]

Step 3: Calculate Committees with 3 Men and 2 Women

For 3 men and 2 women: - Choose 3 men from 8 men: \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. \] - Choose 2 women from 11 women: \[ \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55. \] Multiply the number of ways to choose men and women: \[ 56 \times 55 = 3080. \]

Step 4: Calculate Committees with No More than 3 Women

Possible combinations are: 0 women (5 men), 1 woman (4 men), 2 women (3 men), and 3 women (2 men). - 0 women and 5 men: Already calculated as 56. - 1 woman and 4 men: \[ \binom{11}{1} \times \binom{8}{4} = 11 \times 70 = 770. \]- 2 women and 3 men: Already calculated as 3080. - 3 women and 2 men: \[ \binom{11}{3} \times \binom{8}{2} = 165 \times 28 = 4620. \]Add up all possible combinations: \[ 56 + 770 + 3080 + 4620 = 8526. \]

Step 6: Conclusion and Answer

From the calculations, we have the following answers: (a) All men: 56 ways. (b) All women: 462 ways. (c) 3 men and 2 women: 3080 ways. (d) No more than 3 women: 8526 ways.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and choosing objects. It is fundamental in solving problems where the order of selection does not matter, as seen in the committee formation problem. The key tool used here is the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]This formula helps calculate the number of ways to choose \(r\) items from \(n\) items without considering the order of selection. In the problem, it was used to find different ways to form committees with specific compositions of men and women.

• Choosing 5 men from 8 involves computing \(\binom{8}{5}\), resulting in 56 ways.
• Choosing 5 women from 11 involves \(\binom{11}{5}\), giving 462 ways.
• For mixed gender committees, combinations were multiplied to account for choosing members from each gender separately.

Combinatorics is essential for understanding how different selections and arrangements work without regard to order, making it a crucial aspect of probability and higher mathematics.

Probability

Probability involves calculating the chance or likelihood of an event occurring. Although probability is not directly calculated in this committee selection problem, understanding combinations is a foundational skill for working with probabilities. In scenarios where outcomes need to be calculated, such as forming a committee with specific criteria, the use of combinations comes into play. For instance, if you wanted to calculate the probability of forming a committee with exactly 3 men and 2 women from the given club, you would use:

• The total number of 5-person committees from 19 people: \(\binom{19}{5}\).
• The number of favorable committees (3 men and 2 women): 3080.
• Probability: \(\frac{3080}{\text{Total committees}}\).

Mastering how to apply combinations in probability prepares you for more complex calculations, where understanding the likelihood of occurrences is essential.

Mathematics Education

Mathematics Education emphasizes understanding concepts deeply to apply them effectively in solving real-world problems. Teaching how to use combinations in committee selection problems enhances problem-solving skills and mathematical thinking. Students can take away several key lessons:

• Understanding fundamentals: Knowing when to use combinations instead of permutations.


• Practical application: Applying formulas in different scenarios to reinforce knowledge.


• Progressive exploration: Learning to handle increasingly complex problems step-by-step, building confidence and proficiency in combinatorial reasoning.

By tackling problems like committee selection, students are not only practicing calculation skills but also learning to think critically and make logical decisions in various mathematical situations.

Algebra

Algebra, the branch of mathematics dealing with symbols and the rules for manipulating these symbols, provides the foundation for solving equations and understanding mathematical principles like combinations. In the context of our committee problem:

• The combination formula itself is an algebraic expression, showcasing manipulation of factorial expressions.


• Simplifying factorials involves basic algebraic technique, requiring understanding and rearrangement of terms.


• Calculating combinations for different committee scenarios involves setting up and solving these algebraic expressions swiftly and accurately.

The algebraic structure in combinations demonstrates how algebra operates at the heart of many mathematical concepts, acting as a bridge to applying these concepts in solving practical problems.