Question

Jodi is parking seven different types of vehicles side by side facing the display window at the dealership where she works. a) In how many ways can she park the vehicles? b) In how many ways can she park them so that the pickup truck is next to the hybrid car? c) In how many ways can she park them so that the convertible is not next to the subcompact?

Short answer

a) 5040 b) 1440 c) 3600

Step-by-step solution

Understanding the problem

Jodi needs to park seven different types of vehicles in a line. We need to find out the number of permutations for different scenarios.

Total number of permutations

The total number of ways to arrange 7 different vehicles is given by the factorial of 7, denoted as 7!. Calculate the factorial: 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1.

Calculate 7!

7! = 5040So, there are 5040 ways to park the vehicles.

Vehicles next to each other (pickup truck and hybrid)

Treat the pickup truck and hybrid car as a single unit. This gives us 6 units (5 other vehicles + 1 combined unit).The number of ways to arrange these 6 units is 6!.Within the unit, the pickup truck and hybrid car can be arranged in 2 ways. Therefore, calculate 6! and multiply by 2.

Calculate 6! and multiply by 2

6! = 720Since the pickup truck and hybrid car can be arranged in 2 ways within the unit, the total = 6! * 2 = 720 * 2 = 1440.

Vehicles not next to each other (convertible and subcompact)

First, calculate the total number of ways to arrange all vehicles (Step 2).Then, calculate the number of ways in which the convertible and subcompact are next to each other (similar to Step 4).Subtract the second result from the first to get the number of ways in which the convertible and subcompact are not next to each other.

Calculate the number of ways convertible and subcompact are next to each other

Treat the convertible and subcompact as a single unit. This gives us 6 units.Calculate 6! and multiply by 2, as the convertible and subcompact can be arranged in 2 ways within the unit.

Subtract to find the number of non-adjacent arrangements

Total arrangements = 7! = 5040Adjacent arrangements (convertible and subcompact next to each other) = 6! * 2 = 1440Non-adjacent arrangements = 7! - 6! * 2 = 5040 - 1440 = 3600.

Key concepts

Factorials

To solve many combinatorics problems, we often need to understand factorials. A factorial, denoted as !, represents the product of all positive integers up to that number. For example, the factorial of 7, denoted as 7!, is calculated as follows: \[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \] Factorials are essential for finding the total number of permutations of a set of unique items. In Jodi's exercise, we use 7! to determine the total number of ways she can park the seven vehicles. This calculation yields 5040 permutations, demonstrating how factorials help us understand the vast number of possible arrangements.

Permutations

Permutations refer to the various ways in which a set of items can be arranged in order. Each unique arrangement is a permutation. For instance, in the context of Jodi's problem, she needs to park seven different vehicles in a line. Using factorials, we determine that there are 5040 unique ways (permutations) to arrange these vehicles. This concept extends to partial permutations as well. For example, if we only consider 3 out of the 7 vehicles, we would calculate the permutations of these three while ignoring the remaining vehicles. The formula for this is: \[ P(n, r) = \frac{n!}{(n-r)!} \] where \( n \) is the total number of vehicles and \( r \) carries the number you are choosing to arrange.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and permutations of objects. It helps us solve problems related to counting, arrangement, and selection. In Jodi's parking problem, combinatorics principles are applied to determine the total and specific arrangements of vehicles. We analyze different scenarios, like vehicles parked next to each other or not next to each other, using combinatorial arguments. By identifying groups or units, like treating two vehicles as one unit, combinatorics makes it easier to calculate complex permutations. This approach simplifies otherwise daunting calculations. For more advanced combinatorics, you might encounter binomial coefficients and the Pigeonhole Principle, which help solve a broader range of problems.

Adjacent Arrangements

In certain problems, specific conditions must be met. One common condition in permutation problems involves items parked or placed next to each other (adjacent arrangements). For Jodi's scenario involving the pickup truck and the hybrid car parked next to each other, we treat them as one single unit. This reduces the number of items to six units (the combined truck-hybrid unit plus the five other vehicles). The number of ways to arrange these six units is given by 6!, which equals 720. Since within this combined unit, the truck and hybrid can switch places, we multiply by 2 to account for this. Thus, the total adjacent arrangements are: \[ 6! \times 2 = 720 \times 2 = 1440 \] This method helps simplify solving problems with adjacent conditions.

Non-Adjacent Arrangements

Non-adjacent arrangements focus on scenarios where certain items cannot be placed next to each other. In Jodi's problem, we must calculate how many ways she can park the vehicles such that the convertible is not next to the subcompact. To solve this, we first need the total number of arrangements (7!), which is 5040. Next, we find the number of arrangements where the convertible and subcompact are adjacent, treating them as a single unit. This gives 6 units, and the number of adjacent arrangements is: \[ 6! \times 2 = 1440 \] To find the non-adjacent arrangements, we subtract the adjacent arrangements from the total arrangements: \[ 5040 - 1440 = 3600 \] Thus, there are 3600 ways to park the vehicles such that the convertible and subcompact are not next to each other.