/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 In Exercises \(1-8,\) use the fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-8,\) use the formula for \(_{n} P_{r}\) to evaluate each expression. $$ _{9} P_{4} $$

Short Answer

Expert verified
The value of \(_{9} P_{4} is 3024.

Step by step solution

01

Identify variables from the exercise

From the notation \(_{9} P_{4}, it is clear that \(n = 9\) and \(r = 4\).
02

Calculate Factorials

Calculate the factorials for \(n\) and \(n - r\). \(9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880\), and \(5! = 5 * 4 * 3 * 2 * 1 = 120\).
03

Use Permutation Formula

Now, plug in these values into the permutation formula \(_{n} P_{r} = n! / (n - r)!. Therefore, \(_{9} P_{4} = 362880 / 120 = 3024\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial Notation
Factorial notation, symbolized by an exclamation point (!), is a mathematical concept that represents the product of all positive integers from 1 up to a given number. For any positive integer n, the factorial is denoted as n! and is calculated by multiplying n by every positive integer less than itself. For example, 5! (read as 'five factorial') is calculated as:
\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120.\]
It's important to note that factorial notation is only defined for non-negative integers. A special case to remember is that 0! is defined to be 1, which is a convention used for various mathematical reasons, including the simplification of formulas.

Factorial notation plays a critical role in permutations, combinations, and other areas of combinatorics, as it allows us to count how many ways we can arrange or select items without having to list every possibility.
Permutation Formula
The permutation formula helps determine the number of ways in which you can arrange a subset of items from a larger set, where the order of arrangement is important. Mathematically, for a set with n distinct items, the number of ways to arrange r items is represented by the permutation notation \( _{n}P_{r} \).
This notation is based on the formula:
\[ _{n}P_{r} = \frac{n!}{(n-r)!} \]
In the context of our example, we want to find the number of ways to arrange 4 items from a set of 9. Using the permutation formula, we need to calculate 9! to represent the possible arrangements of all nine items, and then divide by (9-4)! to correct for the arrangements of the remaining items that we're not considering.
Evaluating Permutations
Evaluating permutations becomes straightforward once you understand factorial notation and the permutation formula. It involves calculating the factorial of the number of items in the set (n!) and dividing by the factorial of the difference between the number of items in the set and the items you want to arrange ((n-r)!).
For instance, to find the number of ways to arrange 4 out of 9 items, represented by \( _{9}P_{4} \), we follow these steps:
  • Calculate the factorial of the total number of items: \(9! = 362880\).
  • Calculate the factorial of the number of items not to be arranged: \(5! = 120\), since 9-4=5.
  • Apply the permutation formula: \(_{9}P_{4} = \frac{9!}{5!} = \frac{362880}{120} = 3024\).
Consequently, there are 3024 different ways to arrange 4 items out of a set of 9.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the given binomial coefficient. $$ \left(\begin{array}{c}15 \\\2\end{array}\right) $$

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$ \left(x^{2}+1\right)^{16} $$

What is the common ratio in a geometric sequence?

Follow the outline on the next page to use mathematical induction to prove that $$ \begin{aligned}(a+b)^{n}=\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\\+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned} $$ a. Verify the formula for \(n=1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$ \begin{aligned}&(a+b)^{k+1}=\left(\begin{array}{l}k \\\0\end{array}\right) a^{k+1}+\left[\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)\right] a^{k} b\\\&\begin{array}{l}+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1} \end{array}\end{aligned} $$ e. Use the result of Exercise 74 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\\ r+1\end{array}\right)$$=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\) then \(\left(\begin{array}{l}k \\ 0\end{array}\right)+\left(\begin{array}{l}k \\\ 1\end{array}\right)=\left(\begin{array}{c}k+1 \\\1\end{array}\right)\) and\(\left(\begin{array}{l}k \\ 1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)=\left(\begin{array}{c}k+1 \\ 2\end{array}\right)\) f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)=\left(\begin{array}{c}k+1 \\ 0\end{array}\right) \quad\) (why?) and \(\left(\begin{array}{l}k \\ k\end{array}\right)=\) \(\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (3 x-y)^{5} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.