Combina8ons combina8ons k element subsets

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Unformatted text preview: , {B} , {C} ⎛ 3⎞ number of two-element subsets = ⎜ =3 2⎟ ⎝ ⎠ {A, B} , {A, C} , {B, C} Example: a set with 3 elements ⎛ 3⎞ number of zero-element subsets = ⎜ ⎟ ≡ 1---just the empty set ⎝0⎠ A Set S = B C ⎛ 3⎞ number of one-element subsets = ⎜ ⎟=3 ⎝1⎠ {A} , {B} , {C} ⎛ 3⎞ number of two-element subsets = ⎜ =3 2⎟ ⎝ ⎠ {A, B} , {A, C} , {B, C} ⎛ 3⎞ number of three-element subsets = ⎜ = 1---just the set S = { A, B, C } itself 3⎟ ⎝ ⎠ ⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ total number of subsets = ⎜ +⎜ +⎜ +⎜ = 1 + 3 + 3 + 1 = 8 = 23 ⎝ 0⎟ ⎝ 1⎟ ⎝ 2⎟ ⎝ 3⎟ ⎠ ⎠ ⎠ ⎠ + b (a + b) = (a + ba ⋅(+ )( b) … ⋅ a ) n n times ⎛ n ⎞ n −1 ⎛ n ⎞ n − 2 2 ⎛ n ⎞ 2 n − 2 ⎛ n ⎞ n −1 = a +⎜ a b+⎜ a b + … + ⎜ ⎟a b + ⎜ ⎟ ab + b n ⎟ ⎟ ⎝ n − 1⎠ ⎝ n − 2⎠ ⎝ 2⎠ ⎝ 1⎠ n Coefficient of a k = number of ways we can choose k a' s from the n terms in parentheses ⎛ n⎞ Convention : ⎜ ⎟ = 1 ⎝ 0⎠ ⎛ n⎞ k n− k n ⎛ n⎞ For example, 2 = (1 + 1) = ∑⎜ ⎟ ⋅ 1 ⋅ 1 = ∑⎜ ⎟ k k k = 0⎝ ⎠ k = 0⎝ ⎠ n n n Ilya Pollak Pascal’s triangle Ilya Pollak Pascal’s triangle Ilya Pollak Pascal’s triangle Ilya Pollak Pascal’s triangle: Deriva8on ⎛ n ⎞ ⎛ n⎞ n! n! + ⎜ ⎟+⎜ ⎟= k − 1⎠ ⎝ k ⎠ ( k − 1)!( n − ( k − 1))! k!( n − k )! ⎝ k ⋅ n! ( n − k + 1) ⋅ n! = + k k − ( n − k + 1)! k!( n k( nk + ⋅ 1)! )! ( − ⋅ − 1) = k! ( n − k +1)! ( k + n − k + 1) ⋅ n! k!( n − k + 1)! ( n + 1) ⋅ n! = k!( n − k + 1)! ( n + 1)! = k!( n − k + 1)! ⎛ n + 1⎞ =⎜ ⎟ k⎠ ⎝ = Ilya Pollak...
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This note was uploaded on 09/11/2013 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue University-West Lafayette.

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