Problem 41 - College Algebra

Question

Use the formula for \(_{n} C\), to solve Exercises \(41-48\). An election ballot asks voters to select three city commissioners from a group of six candidates. In how many ways can this be done?

Step-by-Step Solution

Answer

Therefore, there are 20 different ways to select three city commissioners from the group of six candidates.

Step 1: Identification

First, identify the values for \(n\) and \(r\). In this case, \(n=6\) (the total number of candidates), and \(r=3\) (the number of commissioners to be selected).

Step 2: Application

Apply the values of \(n\) and \(r\) to the combinations formula \(_{n}C_{r} = \frac{n!}{r!(n-r)!}.\) Hence, compute \(_{6}C_{3} = \frac{6!}{3!(6-3)!}\).

Step 3: Calculation

Calculate the factorials and solve the equation. The factorial of a number is the product of all positive integers less than or equal to that number. For example, \(6! = 6*5*4*3*2*1\), \(3! = 3*2*1\), and \((6-3)! = 3! = 3*2*1\). So, \(_{6}C_{3} = \frac{6*5*4*3*2*1}{3*2*1*3*2*1} = 20.\)

Key Concepts

Combination Formula

Understanding the combination formula is essential for solving problems involving the selection of items from a larger set where order does not matter. For instance, determining how many ways you can choose a committee, a group for a project, or even lottery numbers.

The combination formula is given by \begin{align*}_{n}C_{r} = \dfrac{n!}{r!(n-r)!}\begin{align*}in which:\

\
• \(n\) represents the total number of items,\
• \(r\) is the number of items to select,\
• \(n!\) denotes the factorial of \(n\),\
• \(r!\) denotes the factorial of \(r\), and\
• \((n-r)!\) is the factorial of the difference between \(n\) and \(r\).\

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To compute the number of combinations, you must first understand factorials, as they are the building blocks of this formula. Factorials are discussed in further detail in the next section.

Factorials

A factorial, represented by an exclamation mark (!), is a function that multiplies a number by all the positive integers below it. For example, the factorial of 5, written as \(5!\), is calculated as:\
\
\begin{align*}5! = 5 \times 4 \times 3 \times 2 \times 1 = 120.\begin{align*}\
Factorials play a crucial role in both permutations and combinations as they are used to figure out the total number of ways items can be arranged (permutations) or selected (combinations). They are especially important when solving for combinations, as seen in the equation for the combination formula mentioned previously.

In the context of the exercise provided, factorials were used to simplify the expression \(_{6}C_{3} = \dfrac{6!}{3! \times (6-3)!}\) by canceling out similar terms in the numerator and the denominator to find the number of ways three city commissioners can be selected from six candidates.

Permutations and Combinations

Permutations and combinations are two foundational concepts in combinatorial mathematics, dealing with the arrangement and selection of a number of objects.

Permutations refer to the arrangements of objects where the order is important. In permutation problems, you are often asked how many different ways you can arrange a certain number of items.

For example

Suppose we have three different books and we want to arrange them on a shelf. The number of different arrangements (permutations) of these three books is given by \(3! = 3 \times 2 \times 1 = 6\).

Combinations, on the other hand, are selections where the order doesn't count. In combination problems, you are usually required to find how many ways you can choose a subset of items from a larger set. Using the combination formula,\begin{align*}_{n}C_{r} = \dfrac{n!}{r!(n-r)!},\begin{align*}we can solve problems like the one in the exercise where we want to know how many ways three city commissioners can be selected from a group of six candidates without worrying about the order of selection.