College Algebra Exam Review 47

# College Algebra Exam Review 47 - ± 1 2 ² : For example,...

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1.9. COUNTING 57 there are just as many subsets of f 1;2;:::;n g as there are sequences of n 0’s and 1’s, namely 2 n . Proposition 1.9.1. A set with n elements has 2 n subsets. How many two–element subsets are there of a set with ﬁve elements? You can list all the two–element subsets of f 1;2;:::;5 g and ﬁnd that there are 10 of them. How many two–element subsets are there of a set with n elements? Let’s denote the number of two–element subsets by ± n 2 ² . A two–element subset of f 1;2;:::;n g either includes n or not. There are n ± 1 two– element subsets that do include n , since there are n ± 1 choices for the second element, and the number of two–element subsets that do not in- clude n is ± n ± 1 2 ² . Thus, we have the recursive relation ± n 2 ² D .n ± 1/ C ± n
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Unformatted text preview: ± 1 2 ² : For example, ± 5 2 ² D 4 C ± 4 2 ² D 4 C 3 C ± 3 2 ² D 4 C 3 C 2 C ± 2 2 ² D 4 C 3 C 2 C 1 D 10: In general, ± n 2 ² D .n ± 1/ C .n ± 2/ C ²²² C 2 C 1: This sum is well known and equal to n.n ± 1/=2 . (You can ﬁnd an inductive proof of the formula .n ± 1/ C .n ± 2/ C ²²² C 2 C 1 D n.n ± 1/=2 in Appendix C.1 .) Here is another argument for the formula ± n 2 ² D n.n ± 1/=2 that is better because it generalizes. Think of building the nŠ permutations of f 1;2;:::;n g in the following way. First choose two elements to be the ﬁrst two (leaving n ± 2 to be the last n ± 2 ). This can be done in ± n 2 ² ways....
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## This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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