ans_prac_final-1

# ans_prac_final-1 - a vertex with degree d Then the sum of...

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CSE 21: Answers to Practice Final 1) (a) g ( n ) = 13 - 1 2 13 ± 3 + 13 2 ! n + 13 + 1 2 13 ± 3 - 13 2 ! n . (b) h ( n ) = ² 1 - 3 4 n ³ 4 n . 2) 1 · ² 5 1 ³ · ² 7 2 ³ ² 12 3 ³ + 2 · ² 5 2 ³² 7 1 ³ ² 12 3 ³ + 3 · ² 5 3 ³ ² 12 3 ³ = 5 4 3) (d) f g h 4) 63. 5) In binomial theorem ( a + b ) n = n X k =0 ² n k ³ a k b n - k , set a = 3 , b = - 1. 6) · ² 13 52 · 39 51 ³ + 2 · ² 13 52 · 12 51 ³ = 21 68 7) (a) 3 5 · 1 3 + 2 5 · 4 7 = 3 7 (b) 3 5 · 1 3 3 5 · 1 3 + 2 5 · 4 7 = 7 15 . 8) 5 · 4 · 3 5 3 = 12 25 9) (a) ² 25 - 4 - 6 + 7 - 1 7 - 1 ³ = 54264 (b) ² 19 + 3 - 1 3 - 1 ³ + ² 4 1 ³² 18 + 3 - 1 3 - 1 ³ + ² 4 2 ³² 17 + 3 - 1 3 - 1 ³ + ² 4 3 ³² 16 + 3 - 1 3 - 1 ³ + ² 4 4 ³² 15 + 3 - 1 3 - 1 ³ = 2744 10) 4 · (2! · 3!) (2 + 3)! = 2 5 11) E = 2 3 · 5 = 10 3 , V ar = " 2 3 - ² 2 3 ³ 2 # · 5 = 10 9 . 12) Prove by contradiction. Suppose there are at most d - 1 vertices of degree 1. Then any of the other n - d +1 vertices has degree at least 2. where n is the total number of vertices. Remember that there is
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Unformatted text preview: a vertex with degree d . Then the sum of degrees is at least 1 Â· ( d-1)+2 Â· ( n-d )+ d = 2 n-1, Since the sum of degrees is twice the number of edges, the number of edges is at least d (2 n-1) / 2 e = n > n-1, a contradiction. 13) (a) 4 7 , 4 7 (b) 2 3 , 1 2 14) This is one of the problems in the last homework. 15) 1 1000 Â· 99 100 1 1000 Â· 99 100 + 999 1000 Â· 1 100 = 11 122 â‰ˆ 9 . 02% 1...
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## This note was uploaded on 04/23/2008 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.

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