a. Explanation Given Information: For k = 0 and S 1 , S 2 , S 3 , ... , S k are mutually exclusive not empty subsets of S such that ∪ i = 1 k S i = S then the set { S 1 , S 2 , S 3 , ... , S n } is a partition of S . T n denotes the number of different partitions of { 1 , 2 , 3 , ... , n } Proof: For T 3 the elements are { 1 , 2 , 3 } . The partitions are: S 1 = { 1 } , { 2 } , { 3 } S 2 = { 1 , 2 } , { 3 } S 3 = { 1 , 3 } , { 2 } S 4 = { 1 } , { 2 , 3 } S 5 = { 1 , 2 , 3 } The number of partitions for T 3 = 5 For T 4 the elements are { 1 , 2 , 3 , 4 } The partitions are: S 1 = { 1 } , { 2 } , { 3 } , { 4 } S 2 = { 1 , 2 } , { 3 , 4 } S 3 = { 1< b. To determine To show : T n + 1 = 1 + ∑ k = 1 n ( n k ) T k

10th Edition

ISBN: 9780134753751

Author: Ross

Publisher: PEARSON

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Set Theory

Every field in the present world involves a set. The theory of sets was developed by German mathematician Georg Cantor while working on the “problems of trigonometric series”.

Question

Chapter 2, Problem 2.8TE

To determine

To show: T3=5 and T4=15 by computing all partitions

To determine

To show: Tn+1=1+∑k=1n( n k)Tk

Expert Solution & Answer

Chapter 2 Solutions

FIRST COURSE IN PROBABILITY (LOOSELEAF)

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T 3 = 5 and T 4 = 15 by computing all partitions

Solution Summary: The author explains that T_3=5 and >=15 by computing all partitions.

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To show: T3=5 and T4=15 by computing all partitions

To show: Tn+1=1+∑k=1n( n k)Tk

Chapter 2 Solutions