Explanation Let S ( n + 1 , m ) denote the number of ways to partition n + 1 elements into m non-empty subsets. Also, let the elements be of the form { 1 , 2 , … , n + 1 } . Now, this n + 1 can remain as it is or it can be grouped with the other elements. Suppose a particular subset contains n + 1 , then there are r other elements in the same partition. Clearly, this can be done in C ( n , r ) S ( n − r , m − 1 ) . This is because, already n + 1 is fixed and the remaining r must be chosen from the n elements and m − 1 more partitions are required from n + 1 − 1 − r = n − r elements. Since to ensure to non-emptiness, at least m − 1 elements are required, that is the value of r varies from 0 ≤ r ≤ n − ( m − 1 ) . Thus, for all integers n > 0 and m > 1 , S ( n + 1 , m ) is evaluated as follows

5th Edition

ISBN: 9780134689562

Author: Dossey, John A.

Publisher: Pearson,

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Statements and Logic

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Chapter 8.6, Problem 32E

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To prove: That for all integers n>0 and m>1, S(n+1,m)=C(n,0)S(n,m−1)+C(n,1)S(n−2,m−1)+⋯+C(n,n−m+1)S(n−1,m−1).

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Chapter 8.6, Problem 31E

Chapter 8.6, Problem 35E

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Chapter 8 Solutions

Discrete Mathematics

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To prove: That for all integers n &gt; 0 and m &gt; 1 , S ( n + 1 , m ) = C ( n , 0 ) S ( n , m − 1 ) + C ( n , 1 ) S ( n − 2 , m − 1 ) + ⋯ + C ( n , n − m + 1 ) S ( n − 1 , m − 1 ) .

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To prove: That for all integers n>0 and m>1, S(n+1,m)=C(n,0)S(n,m−1)+C(n,1)S(n−2,m−1)+⋯+C(n,n−m+1)S(n−1,m−1).

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Chapter 8 Solutions

To prove: That for all integers n > 0 and m > 1 , S ( n + 1 , m ) = C ( n , 0 ) S ( n , m − 1 ) + C ( n , 1 ) S ( n − 2 , m − 1 ) + ⋯ + C ( n , n − m + 1 ) S ( n − 1 , m − 1 ) .