Problem 9
Question
In \(3-22,\) evaluate each expression. $$ _{8} P_{4} $$
Step-by-Step Solution
Verified Answer
The expression \(_{8}P_{4}\) evaluates to 1,680.
1Step 1: Understand the permutation notation
The notation \(_{n}P_{r}\) represents a permutation, which refers to the number of ways to arrange \(r\) objects from a total of \(n\) objects. Permutations account for the order of arrangement.
2Step 2: Recall the permutation formula
The formula for permutations \(_{n}P_{r}\) is given by: \[ _{n}P_{r} = \frac{n!}{(n-r)!}\]where \(n!\) is the factorial of \(n\).
3Step 3: Plug numbers into the formula
Substitute \(n = 8\) and \(r = 4\) into the permutation formula:\[ _{8}P_{4} = \frac{8!}{(8-4)!} = \frac{8!}{4!}\]
4Step 4: Simplify the factorials
Calculate the factorial values:- \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)- \(4! = 4 \times 3 \times 2 \times 1\)Then,\[ _{8}P_{4} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\]
5Step 5: Cancel common factors
Cancel out the common factors \(4!,\) which is \(4 \times 3 \times 2 \times 1\), from the numerator and the denominator:\[ _{8}P_{4} = 8 \times 7 \times 6 \times 5\]
6Step 6: Compute the final result
Multiply the remaining numbers:\[ 8 \times 7 = 56\]\[ 56 \times 6 = 336\]\[ 336 \times 5 = 1680\]The final result is that there are 1,680 ways to arrange 4 objects out of 8.
Key Concepts
Permutation formulaFactorialsOrder of arrangementCombinatorics
Permutation formula
When we talk about permutations, we refer to the number of distinct ways to arrange a set of elements where order is important. The permutation formula allows you to calculate this arrangement for selecting \(r\) items out of \(n\) total items. This is expressed mathematically as:\[ _{n}P_{r} = \frac{n!}{(n-r)!} \]where:
- \(n!\) represents the factorial of \(n\), which is the product of all positive integers up to \(n\).
- \((n-r)!\) refers to the factorial of the difference between \(n\) and \(r\).
Factorials
Factorials are a fundamental concept in permutations, and they are symbolized by an exclamation mark (!). To compute the factorial of a number \(n\), denoted as \(n!\), you multiply \(n\) by every positive integer less than it, down to 1. Here is a breakdown of what factorials look like:
- \(3! = 3 \times 2 \times 1 = 6\)
- \(4! = 4 \times 3 \times 2 \times 1 = 24\)
- \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
Order of arrangement
In permutations, the order of arrangement is crucial. This notion signifies that not only the elements selected are important, but also the sequence in which they are arranged. If the sequence changes, it counts as a different permutation.
Consider, for example, arranging the letters A, B, and C. The different permutations would include ABC, ACB, BAC, BCA, CAB, and CBA. Each permutation is unique because of the different order of letters.
Because order matters, permutations are different from combinations, where only the selection of items is important, not their sequence. This distinction can significantly change how you calculate problems in combinatorics, especially when determining possibility counts.
Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and combination of objects. It is fundamental in analyzing the finite possibilities or arrangements of objects to solve problems related to counting.
One of the core parts of combinatorics is studying permutations to understand arrangements where the order matters. This field also covers combinations, where order does not matter, as well as other concepts like graph theory, partition theory, and more.
Combinatorics helps solve various real-world problems including finding the probability of events, optimizing networks, and analyzing data structures in computer science. By using these principles, you can gain insights into the behavior of mathematical and physical systems, making it an indispensable tool for STEM fields.
Other exercises in this chapter
Problem 9
In \(3-10,\) write the expansion of each binomial. $$ (2 b-1)^{3} $$
View solution Problem 9
Three cards are drawn from a standard deck of 52 cards without replacement. What is the probability that all three cards are kings?
View solution Problem 9
For the given values of \(r\) and \(n,\) find the number of ordered selections of \(r\) objects from a collection of \(n\) objects with replacement. \(r=2, n=8\
View solution Problem 10
In \(3-10,\) write the expansion of each binomial. $$ (1-i)^{5} ; i=\sqrt{-1} $$
View solution