/*! 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 18 Find the value of each permutati... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the value of each permutation. $$ { }_{9} P_{4} $$

Short Answer

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3024

Step by step solution

01

- Understand the permutation notation

The notation _9P_4 denotes the number of ways to arrange 4 objects out of 9 distinct objects. The general formula for permutations is given by _\({n}P_{r} = \frac{n!}{(n-r)!}\).
02

- Identify the values of n and r

For this permutation, \({n} = 9\) and \({r} = 4\). We will use these values in the permutation formula.
03

- Calculate the factorial of n

Calculate \(9!\). \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\). This is the numerator of our formula.
04

- Calculate the factorial of (n-r)

Calculate \((9-4)! = 5!\). \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). This is the denominator of our formula.
05

- Compute the permutation value

Now use the values found in previous steps to compute the permutation: _\({9}P_{4} = \frac{9!}{5!} = \frac{362880}{120} = 3024\). Therefore, _\({9}P_{4} = 3024\).

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Key Concepts

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

Factorial Calculation
The concept of factorial is fundamental in the field of permutations. A factorial of a non-negative integer n, denoted by \(!\), is the product of all positive integers less than or equal to n. For example, \(!4 = 4 \times 3 \times 2 \times 1 = 24\). Factorial calculation helps in counting the arrangements of objects, making it critical for solving permutation problems. In general, \(!n\) is calculated as follows:
\(n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1\).
Let's take another example, \(!5 = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Understanding factorials allows you to grasp more complex concepts, such as permutations and combinations, which involve selecting and arranging objects.
Arrangement of Objects
The arrangement of objects is a key concept in permutations. When we talk about arranging objects, we are interested in understanding how many different ways we can order them. This concept is crucial when solving problems where the order of selection matters. For instance, if we want to find out how many ways we can arrange 4 objects out of 9, we use a permutation.
In permutations, we consider both the selection of the objects and the order in which they are arranged. Suppose we have 3 objects A, B, and C. The different permutations would be ABC, ACB, BAC, BCA, CAB, and CBA, totaling 6 ways. As you can see, changing the order creates a new arrangement, reinforcing why order is important in permutations.
Permutation Formula
The permutation formula is a mathematical way to determine the number of possible arrangements of a subset of objects from a larger set. The general permutation formula is \(_{n}P_{r}\), which represents the number of ways to arrange r objects out of a set of n distinct objects. This formula is given by:
\({n}P_{r} = \frac{n!}{(n-r)!}\).
Here, \(!n\) is the factorial of n, and \(!(n-r)\) is the factorial of the difference between n and r.
For example, to find \(_{9}P_{4}\), we use the values n = 9 and r = 4. Plugging these into our formula, we get:
\(_{9}P_{4} = \frac{9!}{5!} = \frac{362880}{120} = 3024\).
Therefore, there are 3024 different ways to arrange 4 objects out of 9 distinct ones. Understanding this formula allows us to solve a wide variety of permutation problems efficiently.

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Most popular questions from this chapter

You are dealt 5 cards from a standard 52-card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52-card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

Outside a home there is a keypad that will open the garage if the correct four-digit code is entered. (a) How many codes are possible? (b) What is the probability of entering the correct code on the first try, assuming that the owner doesn’t remember the code?

The weather forecast says there is a \(10 \%\) chance of rain on Thursday. Jim wakes up on Thursday and sees overcast skies. Since it has rained for the past three days, he believes that the chance of rain is more likely \(60 \%\) or higher. What method of probability assignment did Jim use?

According to the U.S. National Center for Health Statistics, \(0.15 \%\) of deaths in the United States are 25 - to 34-year-olds whose cause of death is cancer. In addition, \(1.71 \%\) of all those who die are \(25-34\) years old. What is the probability that a randomly selected death is the result of cancer if the individual is known to have been \(25-34\) years old?

In airline applications, failure of a component can result in catastrophe. As a result, many airline components utilize something called triple modular redundancy. This means that a critical component has two backup components that may be utilized should the initial component fail. Suppose a certain critical airline component has a probability of failure of 0.006 and the system that utilizes the component is part of a triple modular redundancy. (a) Assuming each component's failure/success is independent of the others, what is the probability all three components fail, resulting in disaster for the flight? (b) What is the probability at least one of the components does not fail?

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