Problem 3
Question
1–6 Evaluate the expression. $$P(11,4)$$
Step-by-Step Solution
Verified Answer
The value of \( P(11, 4) \) is 7920.
1Step 1: Understand the Permutation Notation
The expression \( P(11, 4) \) represents a permutation, which refers to the number of ways to arrange 4 items from a set of 11 distinct items. The formula for permutations is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \]where \( n \) is the total number of items, and \( r \) is the number of items to arrange.
2Step 2: Apply the Permutation Formula
Apply the formula for permutations to find \( P(11, 4) \). Here, \( n = 11 \) and \( r = 4 \). Substitute these values into the formula: \[ P(11, 4) = \frac{11!}{(11-4)!} = \frac{11!}{7!} \]
3Step 3: Evaluate 11! and 7!
Calculate the factorial of 11 and 7:\[ 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]Since \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \), you only need the top four terms starting from 11: \[ 11 \times 10 \times 9 \times 8 \]
4Step 4: Simplify the Expression
Focus only on the calculation needed based on the factorials. Simplify the fraction:\[ \frac{11!}{7!} = 11 \times 10 \times 9 \times 8 \]Calculate this step-by-step: 1. \( 11 \times 10 = 110 \)2. \( 110 \times 9 = 990 \)3. \( 990 \times 8 = 7920 \)
5Step 5: Conclude the Solution
Conclude the calculation by stating the solution for \( P(11, 4) \). The number of permutations of 4 items from a set of 11 is: \[ P(11, 4) = 7920 \]
Key Concepts
FactorialArrangementCombinatorics
Factorial
The concept of a factorial plays a crucial role in permutations and other combinatorial calculations. In mathematics, a factorial, denoted by an exclamation mark (!), refers to the product of an integer and all the positive integers below it. For instance, the factorial of 5, written as 5!, is calculated as: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \] Factorials are used to determine the total number of ways to arrange a given set of items. When calculating permutations, factorials help in finding out the total number of possible ordered arrangements.
- The factorial of zero, 0!, is defined to be 1. Although it may seem counterintuitive, this definition is useful in mathematical formulas and ensures consistency.
- Factorials grow very quickly and are essential when dealing with large numbers of items. This rapid growth needs to be managed carefully in computational processes.
Arrangement
Arrangement is a key aspect of permutations in combinatorics. When we talk about permutations, we are essentially discussing the different ways to arrange a certain number of items. For example, if you have three books and you want to know in how many ways you can arrange them on a shelf, you would use permutations. Permutations are all about order. The order in which items are arranged matters significantly. This is different from combinations, where order does not matter. Permutations are calculated using the permutation formula: \[ P(n, r) = \frac{n!}{(n-r)!} \]
- Here, \( n \) represents the total number of items.
- \( r \) represents the number of items to be arranged.
Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. It is a field that provides methods to count the number of ways certain patterns can be formed. One of the principal aspects of combinatorics is understanding how items can be selected and arranged in various ways.
In combinatorics, permutations are used when the order of the arrangement matters. This is different from combinations, another concept in combinatorics, where the order of selection does not matter. Understanding this distinction is crucial for solving problems effectively.
- Combinatorics plays a pivotal role in computer science, probability, and statistics. It provides foundational principles for algorithm design and analysis.
- The use of permutations and combinations allows us to solve complex problems related to data arrangement and probability.
Other exercises in this chapter
Problem 3
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