Problem 59 - A Graphical Approach to College Algebra

Question

\text {Solve each problem involving combinations.} Convention Delegation Choices A city council is composed of 5 liberals and 4 conservatives. Three members are to be selected randomly as delegates to a convention. (a) How many delegations are possible? (b) How many delegations could have all liberals? (c) How many delegations could have 2 liberals and 1 conservative? (d) If 1 member of the council serves as mayor, how many delegations are possible that include the mayor?

Step-by-Step Solution

Answer

(a) 84; (b) 10; (c) 40; (d) 28.

Step 1: Calculate Total Delegations

First, identify the total number of council members, which is 9 (5 liberals + 4 conservatives). We need to determine how many ways we can choose 3 members out of these 9. Use the combination formula: \( C(n, k) = \frac{n!}{k!(n-k)!} \). Thus, the number of ways to choose the delegates is \( C(9, 3) = \frac{9!}{3!(9-3)!} = 84 \).

Step 2: Calculate Delegations with All Liberals

We want to determine how many ways we can choose 3 delegates, all of whom are liberals. Use the combination formula on the 5 liberals: \( C(5,3) = \frac{5!}{3!(5-3)!} = 10 \).

Step 3: Calculate Delegations with 2 Liberals and 1 Conservative

First, calculate the number of ways to choose 2 liberals out of 5: \( C(5,2) = \frac{5!}{2!(5-2)!} = 10 \). Next, calculate the number of ways to choose 1 conservative out of 4: \( C(4,1) = \frac{4!}{1!(4-1)!} = 4 \). Multiply these results to find the total number of delegations: \( 10 \times 4 = 40 \).

Step 4: Calculate Delegations Including the Mayor

Assume the mayor is already selected. From the remaining 8 council members, select 2 more members. Calculate using the combination formula: \( C(8,2) = \frac{8!}{2!(8-2)!} = 28 \).

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with the counting of different arrangements. It helps us understand how many ways we can choose or arrange items in a set. In this exercise, when determining the number of possible delegations from a council, combinatorics is the key.

• It allows us to calculate combinations, which are selections where the order doesn't matter.
• We use the combination formula, often denoted as \( C(n, k) = \frac{n!}{k!(n-k)!} \), to find out how many ways we can select \( k \) items from \( n \) items without regard to order.

In our problem, finding delegations involves choosing subsets of council members, emphasizing that order isn't important. Combinatorics provides the tools to solve these particular scenarios.

Permutations

Permutations differ from combinations in that order does matter. While permutations were not directly used in the original problem, understanding the difference is crucial for tackling related problems.

• Permutations are all about arranging items in different sequences. If order mattered in our city council problem, we would use permutations.
• Permutation formula: \( P(n, k) = \frac{n!}{(n-k)!} \), used to find all possible orderings of \( k \) items from a set of \( n \) items.

Know when to use permutations versus combinations is essential. For tasks like selecting delegates where order isn't relevant, combinations suffice, but permutations are vital when sequence matters, like organizing committee roles.

Probability

Probability helps us quantify the likelihood of events. In our scenario, understanding the number of possible delegations forms the basis for any probability calculations involving delegate selection.

• Calculating total possibilities, such as the number of delegations, aids in finding the probability of specific outcomes.
• If each council member has an equal chance of being chosen, probability measures how likely a given outcome, like all liberals, is selected.

While direct probability was not part of the original steps, knowing the total combinations sets up how we can calculate probabilities in similar contexts. For example, finding the probability that two liberals and one conservative are chosen would involve determining the favorable outcomes over the total.

Algebra

Algebra often underlies combinatorial calculations with its equations and variables. In our example, algebra helps simplify the factorial calculations involved in combinations.

• Factorials (e.g., \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \)) are a foundational algebraic concept used in the combination formula.
• Algebra allows us to solve factorial expressions and break down complex calculations into manageable parts.

Understanding how to manipulate algebraic expressions, like those involving factorials seen in our delegate selection problem, provides a toolkit to solve advanced counting problems efficiently.