Chapter 14: Problem 14
Find the value of each permutation. $$ P(8,3) $$
Short Answer
Expert verified
P(8,3) = 336
Step by step solution
01
- Understanding Permutations
Permutations refer to the number of ways to arrange a subset of items from a larger set. The formula for permutations is given by \[ P(n,r) = \frac{n!}{(n-r)!} \] where - \(n\) is the total number of items, - \(r\) is the number of items to be arranged.
02
- Identify Values
Identify the values of \(n\) and \(r\) from the given problem. Here, \(n = 8\) and \(r = 3\).
03
- Apply the Formula
Substitute the values of \(n\) and \(r\) into the permutation formula: \[ P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} \]
04
- Calculate Factorials
Calculate the factorials: \[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \] and \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
05
- Simplify the Expression
Simplify the expression by dividing the factorials: \[ P(8,3) = \frac{8!}{5!} = \frac{40320}{120} = 336 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
factorials
Factorials play a crucial role in permutations and other combinatorial calculations. In mathematics, a factorial is the product of all positive integers up to a given number, denoted as \(n!\). For example, \(5!\) (read as 'five factorial') is calculated as
In the example \(P(8,3)\), we calculate \(8!\) (which equals 40,320) and \(5!\) (which equals 120). Then, we divide these factorials to get the final answer for the permutation:
\[ \frac{8!}{5!} = \frac{40320}{120} = 336 \]
Factorials help break down complex problems into more manageable calculations.
- \(5 \times 4 \times 3 \times 2 \times 1 = 120\)
In the example \(P(8,3)\), we calculate \(8!\) (which equals 40,320) and \(5!\) (which equals 120). Then, we divide these factorials to get the final answer for the permutation:
\[ \frac{8!}{5!} = \frac{40320}{120} = 336 \]
Factorials help break down complex problems into more manageable calculations.
arrangements
The concept of arrangements is essential when discussing permutations. Arrangements refer to the different ways in which a particular set of items can be organized. In permutations, we are specifically interested in the arrangements where the order matters.
If you have three books, you can arrange them in several ways on a shelf. For instance, if the books are labeled A, B, and C, some possible arrangements would be ABC, ACB, BAC, and so forth. Each unique ordering is a different arrangement.
The formula for permutations \[ P(n,r) = \frac{n!}{(n-r)!} \]
helps us calculate how many such arrangements can be made when selecting \(r\) items from \(n\) total items.
As shown in the original exercise, \( P(8,3) \) calculates the number of ways to choose and arrange 3 items out of 8. By following the formula and solving step-by-step, we found there are 336 possible arrangements of 3 items from a set of 8. These arrangements are fundamental in many fields, including logistics, computer science, and everyday decision-making.
If you have three books, you can arrange them in several ways on a shelf. For instance, if the books are labeled A, B, and C, some possible arrangements would be ABC, ACB, BAC, and so forth. Each unique ordering is a different arrangement.
The formula for permutations \[ P(n,r) = \frac{n!}{(n-r)!} \]
helps us calculate how many such arrangements can be made when selecting \(r\) items from \(n\) total items.
As shown in the original exercise, \( P(8,3) \) calculates the number of ways to choose and arrange 3 items out of 8. By following the formula and solving step-by-step, we found there are 336 possible arrangements of 3 items from a set of 8. These arrangements are fundamental in many fields, including logistics, computer science, and everyday decision-making.
combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangements, and combinations. It is widely used in fields such as computer science, statistics, and probability theory.
One of the core concepts in combinatorics is understanding permutations, which is the focus of our exercise. Combinatorial methods help us determine how many ways we can select and organize items from a larger set, whether the order matters (as in permutations) or not (as in combinations).
The formula for permutations provided in the exercise \[ P(n,r) = \frac{n!}{(n-r)!} \]
allows us to compute such quantities easily.
By breaking problems down into simpler, smaller parts using combinatorial methods, we can solve complex real-world problems more effectively.
One of the core concepts in combinatorics is understanding permutations, which is the focus of our exercise. Combinatorial methods help us determine how many ways we can select and organize items from a larger set, whether the order matters (as in permutations) or not (as in combinations).
The formula for permutations provided in the exercise \[ P(n,r) = \frac{n!}{(n-r)!} \]
allows us to compute such quantities easily.
- Suppose you are organizing a contest and need to assign 1st, 2nd, and 3rd place awards to 3 winners from a pool of 8 contestants.
- Using combinatorial concepts like permutations, you can quickly determine there are 336 ways to assign these awards.
By breaking problems down into simpler, smaller parts using combinatorial methods, we can solve complex real-world problems more effectively.