Problem 45 - College Algebra with Modeling and Visualization

Question

(Refer to the discussion after Example \(4 .\) ) A salesperson must travel to 3 of 7 cities. Direct travel is possible between every pair of cities. How many arrangements are there in which the salesperson could visit these 3 cities? Assume that traveling a route in reverse order constitutes a different arrangement.

Step-by-Step Solution

Answer

There are 210 different arrangements.

Step 1: Understanding the Problem

We are given 7 cities and need to find all possible ways to choose and arrange 3 out of these cities for a salesperson's visit. Carefully note that traveling any route in reverse is counted as a different arrangement.

Step 2: Combinatorial Selection

First, determine how many ways we can select 3 cities out of 7. This is a combination problem because the order of selection does not matter initially. We use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]where \( n = 7 \) and \( r = 3 \). Substitute these values and compute.

Step 3: Calculate Number of Combinations

Substituting the values, we have:\[ \binom{7}{3} = \frac{7!}{3! \cdot (7-3)!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 \]This means there are 35 different ways to choose 3 cities out of 7.

Step 4: Permutations of the Selected Cities

Once 3 cities are selected, we need to find the number of ways to arrange these 3 cities, as the order matters. This is a permutation problem within the selected 3 cities. We use the permutation formula: \[ 3! = 3 \cdot 2 \cdot 1 = 6 \]So, there are 6 arrangements for each selection of 3 cities.

Step 5: Total Arrangements

Finally, multiply the number of combinations by the number of permutations to find the total number of arrangements: \[ 35 \times 6 = 210 \]Thus, there are 210 different ways to arrange visits to 3 out of the 7 cities, considering each reverse route as unique.

Key Concepts

Combinatorial Selection

When working on problems involving selection from a group, it's important to understand the concept of combinatorial selection. This means choosing a subset from a larger set where the order of selection does not initially matter. To solve these problems, we use combinations, which are calculated using the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Here, \( n \) represents the total number of items you can choose from, and \( r \) is the number of items to select. Factorials (\(!\)) are used in this formula, meaning you multiply all whole numbers from the number down to 1.

• \( n! \) is the factorial of the total number of items.
• \( r! \) is the factorial of the number of items being chosen.

In the example with the 7 cities, choosing 3 cities involves calculating \( \binom{7}{3} \). This value tells us how many different groups of 3 cities can be selected. Once the cities are selected, the next step is to arrange them.

City Travel Arrangements

City travel arrangements require not only choosing the cities but also deciding on the sequence in which they are visited. In our example, once 3 cities are chosen from a possible 7, the order or sequence in which these cities are visited matters because traveling in reverse is considered a different route. This means each different selection of cities must be further arranged, a task that is typically solved using permutations. Thus, the travel scenario expands from just picking cities, into planning the exact path of travel.

• Select 3 cities — a combinatorial selection problem.
• Arrange these selections — transitioning to a permutation problem.

Each arrangement provides a unique travel plan, greatly increasing possible travel routes. Consider how this concept applies in real-life city travel; each route can offer new roads, landmarks, and potential experiences.

Permutation Formula

Once you've selected your set of items or locations, permuting them involves arranging the items in every possible order. This concept assumes importance in many scenarios, including our city travel case. Now that 3 cities are selected, permutations help calculate the number of ways these cities can be ordered. The formula used here is: \[ r! \] For our problem, it translates as 3!, or \( 3 \times 2 \times 1 = 6 \). This tells us there are 6 different sequences in which any selected trio of cities can be visited.

• An important aspect of permutation is considering the order of items.
• Permutations differ from combinations in that order is critical.

Finally, by multiplying the number of combinations by the permutations of each combination, you find the total possible arrangements. This reflects arrangements where sequence matters, leading us to a total of 210 distinct city travel plans in this scenario.