Problem 36 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In how many ways can the names of six candidates for political office be listed on a ballot?

Step-by-Step Solution

Answer

There are 720 distinct ways to arrange the six candidates for political office on a ballot.

Step 1: Understand Permutations

A permutation is an arrangement of objects in a specific order. In this problem, we have 6 candidates, and we need to arrange them in a specific order. Since the order matters, we'll be using the concept of permutations to solve this problem.

Step 2: Number of ways to arrange 6 candidates

To find out the number of distinct ways to arrange these candidates on a ballot, we can use the permutation formula, which is defined as: \(nPr = \frac{n!}{(n-r)!}\) Where: - \(n\) is the total number of candidates - \(r\) is the number of candidates to be arranged on the ballot - \(nPr\) is the number of distinct permutations - \(!\) denotes the factorial, which is the product of all positive integers less than or equal to the given number. In our case, we have 6 candidates, and we want to find out the number of different arrangements of these 6 candidates. So, \(n = 6\) and \(r = 6\)

Step 3: Plug in and Solve

Now, we'll plug in the given values to the permutation formula: \(6P6 = \frac{6!}{(6-6)!}\) Calculate the factorials \(6P6 = \frac{6!}{0!}\) Since \(0!\) is equal to 1, we have: \(6P6 = 6!\) Now let's find the factorial of 6: \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) So, there are 720 distinct ways to arrange the six candidates for political office on a ballot.

Key Concepts

Factorial

When we look at the factorial, denoted by an exclamation mark \textbf{(!)}, we are referring to a fundamental principle in mathematics that is widely used in permutations and combinations. A factorial represents the product of an integer and all the non-zero integers below it. For example, the factorial of 6, symbolized as \textbf{6!}, is calculated by multiplying 6 by every positive integer less than 6.

To give a clear picture, \textbf{6!} is equal to: \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\textbf{}\).
What often puzzles students is the special case of \textbf{0!}, which is defined to be 1. This might seem counterintuitive at first, but it is essential for many combinatorial formulas to work correctly. Remember that in any permutation problem, you'll need to be comfortable with finding factorials, as they are key in determining the number of ways to arrange or select items.

Arrangement of Objects

Discussing arrangements plays a crucial role in understanding permutations. The heart of permutation is the arrangement of a set number of objects in a specific order. This concept comes into play when the sequence of those objects matters. Let's envision a simple situation: arranging books on a shelf. If you have three books and you want to examine all the possible ways they can be ordered, the arrangement is your focus.

Each different sequence represents a distinct arrangement, and therefore, a separate permutation. For instance, a shelf holding a math textbook, a novel, and a dictionary can have several arrangements like Math-Novel-Dictionary, Novel-Math-Dictionary, and so on. In permutations, as is the case with listing candidates on a ballot, each possible order is unique, which could influence the outcome—in this case, perhaps influencing a voter's choice.

Permutation Formula

The permutation formula is an elegant piece of mathematical shorthand that simplifies the process of determining the number of possible arrangements of objects. Represented as \textbf{nPr}, the formula is \(nPr = \frac{n!}{(n-r)!}\),
where \textbf{n} is the total number of objects, and \textbf{r} is the number of objects we want to arrange.

Using the Permutation Formula

To utilize this formula, you'll start by identifying the value of \textbf{n}, which is the total number of objects you have. Next, decide on \textbf{r}, the specific number of objects you want to arrange from \textbf{n}. By plugging these values into the formula and calculating the factorials, you will determine the total number of distinct permutations.

In educational exercises like ours with the ballot listing, where you're arranging 6 candidates in 6 spots, you would set both \textbf{n} and \textbf{r} to 6. Simplifying the formula, you quickly find that the number of different permutations equals \textbf{6!}, which gives you 720 possible ways those candidates can appear on the ballot. This formula not only aids in answering such textbook questions but also enriches your understanding of how ordered sets can be manipulated.