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1 Melikyan/DM/Fall09 Discrete Mathematics Ch. 6 Counting and Probability Instructor: Hayk Melikyan melikyan@nccu.edu Today we will review sections 6.6, 6.7
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2 Melikyan/DM/Fall09 Algebra of Combinations and Pascal’s Triangle The number of r-combinations from a set of n elements equals the number of (n – r) combinations from the same set. C(n, r) = C(n, n-r)
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3 Melikyan/DM/Fall09 Pascal's Triangle: Theorem: (Pascal’s Formula). Let n and r be positive integers and r n, then C(n + 1, r) = C(n, r – 1) + C(n, r) Proof: Algebraic verses Combinatorial A combinatorial proof is an argument that establishes an algebraic fact by relying on counting principles.
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4 Melikyan/DM/Fall09 Proof (Combinatorial Version) Is S is an set with n+1 elements. The number of all subsets of size r can be calculated as follows:
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5 Melikyan/DM/Fall09 Pascal’s Triangle
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6 Melikyan/DM/Fall09 Exercises Show that: 1 * 2 + 2 * 3 + n * (n + 1) = 2 * C(n + 2, 3) Prove that : C(n, 0) 2 + C(n, 1) 2 + … + C(n, n) 2 = C(2n, n)
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7 Melikyan/DM/Fall09 Binomial Formula Theorem (Binomial Theorem): Given any real numbers a and b and any nonnegative integer n (a + b) n = = a n + C(n, 1)a n-1 b + C(n, 2)a n-2 b 2 +... + C(n, n-1)a n-1 b + b n = =
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8 Melikyan/DM/Fall09 Example: 1. Show that = 2 n 2. Show that = 0 3. Express in the closed form
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9 Melikyan/DM/Fall09 More Combinatorial Proof Let S be all n-card hands that can be dealt from a deck containing n red cards (numbered 1,..., n) and 2n black cards (numbered 1,..., 2n). The right hand side = # of ways to choose n cards from these 3n cards. The left hand side = # of ways to choose r cards from red cards x # of ways to choose n-r cards from black cards = # of ways to choose n cards from these 3n cards = the right hand side.
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10 Melikyan/DM/Fall09 What about ?
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