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Ch. 8 - Sequences, Induction, and Probability
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 9, Problem 5

Use the formula for nPr to evaluate each expression. 6P6

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1
Recall the formula for permutations: \(nP_r = \frac{n!}{(n-r)!}\), where \(n\) is the total number of items and \(r\) is the number of items to arrange.
Identify the values of \(n\) and \(r\) in the expression \$6P6\(. Here, \)n = 6\( and \)r = 6$.
Substitute these values into the formula: \(6P6 = \frac{6!}{(6-6)!}\).
Simplify the denominator: \((6-6)! = 0!\). Remember that by definition, \$0! = 1$.
The expression now becomes \(6P6 = \frac{6!}{1} = 6!\). To find the value, you would calculate \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\), but the problem only asks for the setup.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Permutation Formula (nPr)

The permutation formula nPr calculates the number of ways to arrange r objects from a set of n distinct objects, where order matters. It is given by nPr = n! / (n - r)!, where '!' denotes factorial.
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Factorial Notation

Factorial, denoted by n!, is the product of all positive integers from 1 up to n. For example, 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Factorials are essential in permutations and combinations calculations.
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Evaluating nPr When r = n

When the number of objects chosen r equals the total number n, nPr simplifies to n!, because (n - n)! = 0! = 1. Thus, nPn = n!, representing all possible arrangements of the entire set.
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