College Algebra Exam Review 55 - 65 1.9 COUNTING This sum can be simpliﬁed as follows ÂÃ n X kn Dn D D 1.n k k/Š D nŠ k D0 Since 1 X k D0 1/k n X

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Unformatted text preview: 65 1.9. COUNTING This sum can be simpliﬁed as follows: ÂÃ n X kn Dn D D . 1/ .n k k /Š D nŠ k D0 Since 1 X k D0 . 1/k n X . 1/k k D0 1 : kŠ 1 is an alternating series with limit 1=e , we have kŠ j1=e n X . 1/k k D0 1 j Ä 1=.n C 1/Š; kŠ so jDn nŠ=e j Ä nŠ=.n C 1/Š D 1=.n C 1/: Therefore, Dn is the integer closest to nŠ=e . Example 1.9.14. Ten diners leave coats in the wardrobe of a restaurant. In how many ways can the coats be returned so that no customer gets his own coat back? The number of ways in which the coats can be returned, each to the wrong customer, is the number of derangements of 10 objects, D10 D 1; 333; 961. The primary goal of our discussion of inclusion-exclusion is to obtain a formula for the Euler ' -function: Deﬁnition 1.9.15. For each natural number n, '.n/ is deﬁned to be the the cardinality of the set of natural numbers k < n such that k is relatively prime to n. Lemma 1.9.16. Let k and n be be natural numbers, with k dividing n. The number of natural numbers j Ä n such that k divides j is n=k . Proof. Say kd D n. The set of natural numbers that are no greater than n and divisible by k is fk; 2k; 3k; : : : ; d k D ng, so the size of this set is d D n=k . I Corollary 1.9.17. If p is a prime number, then for all k p k 1 .p 1/. 1, '.p k / D ...
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