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Evaluate the permutations. $$ P_{1}^{20} $$

Short Answer

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Question: Evaluate the permutation \(P_{1}^{20}\). Answer: The evaluated permutation \(P_{1}^{20}\) is equal to 20.

Step by step solution

01

Identify n and r

In the given permutation, \(P_{1}^{20}\), we have \(n=20\) and \(r=1\).
02

Apply the permutation formula

Using the formula for permutations, we have: $$ P_{r}^{n} = \frac{n!}{(n-r)!} $$ Substitute \(n=20\) and \(r=1\) into the formula: $$ P_{1}^{20} = \frac{20!}{(20-1)!} $$
03

Simplify

Simplify the expression by calculating the factorials and subtraction: $$ P_{1}^{20} = \frac{20!}{(20-1)!} = \frac{20!}{19!} $$
04

Calculate the result

Now, use the property of factorials to cancel out common terms: $$ P_{1}^{20} = \frac{20!}{19!} = \frac{20\times19\times18\times \dots \times 3\times 2\times 1}{19\times18\times \dots \times 3\times 2\times 1} $$ Cancelling out the common terms: $$ P_{1}^{20} = 20 $$ The evaluated permutation \(P_{1}^{20}\) is equal to 20.

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Key Concepts

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

Factorials
Factorials are a fundamental mathematical concept used in permutations to count the number of ways to arrange a set of items. The notation for a factorial is "!". For example, the factorial of a number, say 5, is written as 5!. This means you multiply all whole numbers from the number down to 1:
  • 5! = 5 × 4 × 3 × 2 × 1
This product results in 120. Factorials grow very quickly, as each value is multiplied by one higher number. For instance, 6! is 720 as you multiply 6 by all the numbers before it.
Factorials are particularly useful in calculating permutations and combinations because they help establish the total number of sequences possible, even when the sequences are quite large. Remember, 0! is defined as 1 since there's only one way to arrange zero objects.
Permutation Formula
Permutations are a specific way to arrange items where the order of arrangement is significant. The formula for permutations is given by:\[P_r^n = \frac{n!}{(n-r)!}\]Here, \(n\) is the total number of items you can choose from, and \(r\) is the number of items to arrange. The expression \(n!/(n-r)!\) represents all the unique sequences of \(r\) items from \(n\) possibilities.
In our example, \(P_1^{20}\), we calculate the permutation of 1 item selected from a set of 20.
  • First, multiply all integers from 20 down to 1 to get 20!, then divide this by 19! (which consists of all integers from 19 down to 1).
This division effectively cancels out all terms except for 20, leaving us with the answer of 20. It shows that there's really only one possible outcome when choosing one distinct item from a group, as shown by the remaining factorial calculation.
Mathematical Evaluation
Mathematical evaluation involves carrying out all the necessary calculations to make sense of a formula and obtain a specific result. In permutations, evaluating the formula involves simplifying the expression through proper use of factorials and cancellation.
For example, simplifying \(\frac{20!}{19!}\), allows us to cancel all common terms in the numerator and the denominator except for what is left as the simplest form:
  • 20 alone, which simplifies from the factorial sequence.
The logic behind simplification relies on knowing how factorials unfold, with each number descending by one till 1. In our example, simplifying the initial setup results in an easily understandable format: choosing one item from 20 options ends with only one valid sequence - picking the specific item itself, giving a result of 20.

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Most popular questions from this chapter

A woman brought a complaint of gender discrimination to an eight-member HR committee. The committee, composed of five females and three males, voted \(5-3\) in favor of the woman, the five females voting for the woman and the three males against. Has the board been affected by gender bias? That is, if the vote in favor of the woman was \(5-3\) and the board members were not biased by gender, what is the probability that the vote would split along gender lines (five females for, three males against)?

The failure rate for a guided missile control system is 1 in \(1000 .\) Suppose that a duplicate, but completely independent, control system is installed in each missile so that, if the first fails, the second can take over. The reliability of a missile is the probability that it does not fail. What is the reliability of the modified missile?

A large number of adults are classified according to whether they were judged to need eyeglasses for reading and whether they actually used eyeglasses when reading. The proportions falling into the four categories are shown in the table. A single adult is selected from this group. Find the probabilities given here. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Used Eyeglasses } \\ \text { for Reading } \end{array} & \\ \hline \text { Judged to Need Eyeglasses } & \text { Yes } & \text { No } \\ \hline \text { Yes } & .44 & .14 \\ \text { No } & .02 & .40 \end{array} $$ a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not. d. An adult used glasses when they didn't need them.

A sample is selected from one of two populations, \(S_{1}\) and \(S_{2},\) with \(P\left(S_{1}\right)=.7\) and \(P\left(S_{2}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}\) or \(S\), has occurred are $$ P\left(A \mid S_{1}\right)=.2 \text { and } P\left(A \mid S_{2}\right)=.3 $$ Use this information to answer the questions in Exercises \(1-3 .\) Use Bayes' Rule to find \(P\left(S_{2} \mid A\right)\).

When an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2}\) and \(S_{3}\), can occur, with \(P\left(S_{1}\right)=.2, P\left(S_{2}\right)=.5,\) and \(P\left(S_{3}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}, S_{2}\), or \(S_{3}\) has occurred are $$ P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3 $$ If event A is observed, use this information to find the probabilities in Exercises 4 -6. \(P\left(S_{2} \mid A\right)\)

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