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How many derangements are possible of the set {1,2,3,4} O 6 O 7 O 8

How many derangements are possible of the set {1,2,3,4} O 6 O 7 O 8

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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**Question:** How many derangements are possible for the set {1, 2, 3, 4}? **Options:** - ○ 6 - ○ 7 - ○ 8 - ● 9 **Explanation:** In combinatorics, a derangement is a permutation of the elements of a set, such that no element appears in its original position. For the given set {1, 2, 3, 4}, the number of derangements can be calculated using the formula for derangements: \[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \] Applying this formula for \( n = 4 \): \[ !4 = 4! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) \] \[ = 24 \left( 1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} \right) \] \[ = 24 \left( 0.5 - 0.1667 + 0.0417 \right) \] \[ = 24 \left( 0.375 \right) \] \[ = 9 \] Thus, the correct answer is 9 derangements.
Transcribed Image Text:**Question:** How many derangements are possible for the set {1, 2, 3, 4}? **Options:** - ○ 6 - ○ 7 - ○ 8 - ● 9 **Explanation:** In combinatorics, a derangement is a permutation of the elements of a set, such that no element appears in its original position. For the given set {1, 2, 3, 4}, the number of derangements can be calculated using the formula for derangements: \[ !n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \] Applying this formula for \( n = 4 \): \[ !4 = 4! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) \] \[ = 24 \left( 1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} \right) \] \[ = 24 \left( 0.5 - 0.1667 + 0.0417 \right) \] \[ = 24 \left( 0.375 \right) \] \[ = 9 \] Thus, the correct answer is 9 derangements.
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