Unformatted text preview: X are mutually exclusive and exhaustive (one must occur), the total number of ways of selecting k objects from n is the sum (~=~) + (nkl), as desired. (b) Use the identity in (a) twice to obtain (~) = (~=D + (n;1) = [(~=D + (~=DJ + [(~=~) + (n;2)J = (~=D +2(~=D + (n;2). Thus (~) (n; 2) _ (~=~) = 2(~ =~) is even. 23. This is just t (~) (3 nk )( _1)k = (31)n = 2n. (Note that a different expansion for 2 n was found k=O in Problem 34.) 24. If n = 1, m = 2 is even. If n = 2, @ = 6 is even, so assume n 2: 3. Then e:) = en; 1) + e:~11) = (2n 2) + (2n 2) + (2n 2) + (2n 1) n n1 n1 n2 = (2n 2) + (2n 2) + 2 (2n 2) n n2 nl so it suffices to prove that (2nn2) + (2:=22) is even, which is the case because these numbers are equal: ( 2n2) = (2n  2)! = (2n2) n n!(n  2)! n2'...
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 Summer '10
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 Graph Theory, Natural number, @, 12k, 4k, 6k

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