EBK FIRST COURSE IN PROBABILITY, A
EBK FIRST COURSE IN PROBABILITY, A
10th Edition
ISBN: 9780134753683
Author: Ross
Publisher: VST
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Textbook Question
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Chapter 1, Problem 1.12TE

Consider the following combinatorial identity: k = 0 n k ( n k ) = n 2 n 1

a. Present a combinatorial argument for this identity by considering a set of n people and determining, in two ways, the number of possible selections of a committee of any size and a chairperson for the committee.

Hint:

i. How many possible selections are there of a committee of size k and its chairperson?

ii. How many possible selections are there of a chairperson and the other committee members?

b. Verify the following identity for n = 1 , 2 , 3 , 4 , 5 : k = 1 n k ( n k ) = 2 n 2 n ( n + 1 )

For a combinatorial proof of the preceding, consider a set of n people and argue that both sides of the identity represent the number of different selections of a committee, its chairperson, and its secretary (possibly the same as the chairperson).

Hint:

  1. How many different selections result in the committee containing exactly k people?
  2. How many different selections are there in which the chairperson and the secretary are the same?
  3. How many different selections result in the chairperson and the secretary being different?

c. Now argue that k = 1 n ( n k ) k 3 = 2 n 3 n 2 ( n + 3 )

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Chapter 1 Solutions

EBK FIRST COURSE IN PROBABILITY, A

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