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109_sp2011_ho_counting_sol

# 109_sp2011_ho_counting_sol - Prove that R is an infnite set...

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109 Spring 2011 - Counting Exercise. A true-false test has 10 questions. (a) Inhowmanywayscanastudentanswerthequestions if every question is answered? Some subset of the 10 questions will get the answer “true” and the rest will be answered “false”. Therefore, it is suﬃcient to count the number of subsets of the 10 questions. Or, equivalently, the number of function f : N 10 →{ “true” , “false” } .Th i s is 2 10 = 1024. (b) Inhowmanywayscanastudentanswerthequestions if there is a penalty for guessing and the student leaves some (or all) answers blank? In this case, we are counting the number of functions f : N 10 →{ “true”, “false”, “blank” } . This is 3 10 = 59049. Exercise. Prove that if n Z + then ( 2 n n ) is even. Proof. Recall the recursive formula for the binomial coeﬃcients: ± 2 n n ² = ± 2 n 1 n ² + ± 2 n 1 n 1 ² . But, (2 n 1) n = n 1, and recall that ( k r ) = ( k k - r ) for all k Z + and 0 r k . Therefore, ± 2 n n ² = ± 2 n 1 n ² + ± 2 n 1 n 1 ² =2 ± 2 n 1 n 1 ² and since ( 2 n - 1 n - 1 ) Z

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Unformatted text preview: Prove that R is an infnite set. Proof. Suppose For a contradiction that R is fnite. By defnition, that means it is either the empty set or has a fnite cardinality. Since 1 ∈ R , R 6 = ∅ . ThereFore, there is n ∈ Z + such that | R | = n . Equivalently, there is a Function f : N n → R with is a bijection. Since bijections have inverses, there is a Function f-1 : R → N n which is a bijection, hence an injection. Notice that N n +17 ⊂ R . ThereFore, we can restrict f to N n +17 to get a Function f ± N n +17 : N n +17 → N n which is an injection (since the restriction oF an injection is an injection). Lemma 10.1.4 says that iF there is an injection From N m to N n then m ≤ n . In our case, m = n + 17 and we conclude that n + 17 ≤ n , a contradiction. ² Note: in the above, the number 17 was chosen arbitrarily; any positive number would work....
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109_sp2011_ho_counting_sol - Prove that R is an infnite set...

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