Chapter 1. Counting

Chapter 1. Counting - § 1 Counting § 1.1 Basic principle...

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Unformatted text preview: § 1 Counting § 1.1 Basic principle 1.1.1 A simple rule: total k jobs n 1 ways to do job 1 n 2 ways to do job 2 . . . n k ways to do job k                ⇒ ( n 1 n 2 ··· n k ) ways to do the k jobs § 1.2 Selection of distinguishable objects 1.2.1 Definition. For any integer n ≥ 0, define the factorial of n , written n !, by n ! = ( 1 , n = 0 , n ( n- 1) ··· 1 , n ≥ 1 . 1.2.2 Theorem. (Binomial Series Theorem) For any real number α and for any | x | < 1, (1 + x ) α = 1 + αx + 1 2! α ( α- 1) x 2 + 1 3! α ( α- 1)( α- 2) x 3 + ··· . In particular, if α = n is a positive integer, we have (1 + x ) n = 1 + n 1 ¶ x + n 2 ¶ x 2 + ··· + n n ¶ x n , where n r ¶ = n ! r !( n- r )! is known as a binomial coefficient . 1.2.3 Ordered selection without replacement Problem : n distinguishable objects, choose r from them without replacement in an ordered sequence. Example : Beauty Contest, 15 candidates, how many ways to award Winner, 1st and 2nd runners-up? Answer: 15 |{z} W inner × 14 |{z} 1 st × 13 |{z} 2 nd 3 General case: n ( n- 1) ··· ( n- r + 1) = n ! ( n- r )! . Equivalent problem : allocate r distinguishable objects into n distinguishable cells, with no cell containing > 1 objects. 1.2.4 Unordered selection without replacement Problem : n distinguishable objects, choose r from them without replacement in an unordered sequence. Example : Beauty Contest, 15 candidates, how many ways to select 5 to enter final round? Answer: 15 5 ¶ General case: Consider a product of n terms (1 + x )(1 + x ) ··· (1 + x ) = (1 + x ) n = 1 + n 1 ¶ x + n 2 ¶ x 2 + ··· + n n ¶ x n ....
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This note was uploaded on 04/29/2010 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

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Chapter 1. Counting - § 1 Counting § 1.1 Basic principle...

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