Problem 21
Question
19–32 These problems involve permutations. Seating Arrangements In how many different ways can six of ten people be seated in a row of six chairs?
Step-by-Step Solution
Verified Answer
There are 151,200 different ways to seat six of ten people in a row.
1Step 1: Understand the Problem
We need to find the number of ways to seat six people from a group of ten in a row of six chairs. This is a permutation problem because the order in which the people are seated matters.
2Step 2: Use the Permutation Formula
The permutation formula to calculate the number of ways to choose and arrange k items from n items is given by: \[ P(n, k) = \frac{n!}{(n-k)!} \]where \(!\) denotes a factorial, which is the product of all positive integers up to that number.
3Step 3: Plug in the Values
In our case, we have \(n = 10\) people and we are choosing \(k = 6\) to arrange them. Substitute these values into the permutation formula: \[ P(10, 6) = \frac{10!}{(10-6)!} = \frac{10!}{4!} \]
4Step 4: Calculate the Factorials
Calculate the factorials:- \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)- \(4! = 4 \times 3 \times 2 \times 1\) Simplify by canceling out \(4!\) in the numerator and denominator.
5Step 5: Simplify the Expression
After canceling \(4!\), we are left with:\[ 10 \times 9 \times 8 \times 7 \times 6 \times 5 \]Now calculate this product.
6Step 6: Final Calculation
Multiply the numbers:\[ 10 \times 9 = 90 \]\[ 90 \times 8 = 720 \]\[ 720 \times 7 = 5040 \]\[ 5040 \times 6 = 30240 \]\[ 30240 \times 5 = 151200 \]
7Step 7: Conclusion of Permutations
The total number of ways to seat six of ten people in a row of six chairs is 151200.
Key Concepts
FactorialsPermutation FormulaCombinatorics
Factorials
Factorials are essential in permutations, as they provide a way to count and arrange items in an ordered sequence. A factorial is represented by an exclamation point, \(n!\), which means multiplying all whole numbers from the chosen number down to 1. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials are used in the permutation formula to calculate the number of different arrangements possible. The value decreases rapidly; hence, they are perfect for problems involving large amounts of data, like ordering or arranging items. Calculating a factorial is straightforward, but remember:
Factorials are used in the permutation formula to calculate the number of different arrangements possible. The value decreases rapidly; hence, they are perfect for problems involving large amounts of data, like ordering or arranging items. Calculating a factorial is straightforward, but remember:
- 0! is always 1.
- Factorials grow very quickly.
- They are the backbone of many statistical formulas.
Permutation Formula
The permutation formula is a critical tool for determining the number of possible arrangements when the order matters. The generic form of a permutation formula is \(P(n, k) = \frac{n!}{(n-k)!}\). In this formula:
- \(n\) stands for the total number of items available.
- \(k\) represents the number of items to arrange.
- The "!" indicates factorial, highlighting how we multiply items in sequence.
Combinatorics
Combinatorics is a branch of mathematics focusing on counting, arranging, and combination-making of sets in various ways. It's a pivotal area in mathematics because it introduces powerful tools and techniques to solve complex problems involving arrangements and selections.
In combinatorics, permutations and combinations are two primary tools. Permutations are about arranging elements where order matters, while combinations deal with selections where order does not matter. Hence:
In combinatorics, permutations and combinations are two primary tools. Permutations are about arranging elements where order matters, while combinations deal with selections where order does not matter. Hence:
- Permutations focus when order is crucial, like assigning seats or roles;
- Combinations are used when order doesn’t hold significance, such as forming groups.
Other exercises in this chapter
Problem 20
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