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

1.9. COUNTING 65 This sum can be simplified as follows: D n D D n X k D 0 . 1/ k n k .n k/Š D nŠ n X k D 0 . 1/ k 1 kŠ : Since 1 X k D 0 . 1/ k 1 kŠ is an alternating series with limit 1=e , we have j 1=e n X k D 0 . 1/ k 1 kŠ j 1=.n C 1/Š; so j D n nŠ=e j nŠ=.n C 1/Š D 1=.n C 1/: Therefore, D n 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, D 10 D 1; 333; 961 . The primary goal of our discussion of inclusion-exclusion is to obtain a formula for the Euler ' -function