This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 10A: PERMUTATIONS and COMBINATIONS
Example 1(a): Example 1(a) Determine the different PERMUTATIONS of three of the first four letters of the alphabet a, b, c, d and list them all. Solution: PERMUTATIONSOrder is important. abc acb bac bca cab cba abd adb bad bda dab dba acd adc cad cda dac dca bcd bdc cbd cdb dbc dcb RULE:
nP = r n! ( n −r )! 4! P= =4! =4 × ×2 × =24 3 1 4 3 There are ways ( 4 −3)! Example 1 (b) If we are NOT concerned with the order then in how many ways can the selection be made? how
Solution: [abc acb bac bca cab cba] {abd adb bad bda dab dba} (acd adc cad cda dac dca) <bcd bdc cbd cdb dbc dcb> COMBINATIONS  If we are NOT concerned with the order
then: RULE: n! nP =r n Cr = (n − r )!r! r!
4 C3 = Here, there are only ! 3 2× ( ) = 3!(44− 3)! = 4 ×× × ×1 1 = 4 32
4 3 The committees are: a b c, a b d, a c d, bcd Example 2 (i): In how many ways can 8 different books be arranged on a shelf? Solution: Is the order important? Yes. Just ask a librarian!!! Solution: n! nP = r (n − r )! 8! 8! = = 8! 8P = 8 (8 − 8)! 0! There are 8! = 40,320 ways of arranging 8 different books on a shelf. Example 2(ii): Find the number of ways in which 4 of 10 teams can be ranked 1st, 2nd, 3rd and 4th? Example 2(ii): Solution: 10 x 9 x 8 x 7 ___ ___ ___ ___ 1st 2nd 3rd 4th
n! nP = r (n − r )! 10 P = 4 10! 10! 10 × 9 × 8 × 7 × 6 × .... ×1 = = = 10 × 9 × 8 × 7 (10 − 4)! 6! 6 × .... ×1 There are 5,040 ways of ranking teams 1st, 2nd, 3rd and 4th. Example 2(iii): The five finalists of a beauty contest are Miss Argentina, Miss Belgium, Miss China, Miss Denmark, and Miss Egypt. In how many ways can the judges choose the champion, the first runnerup and the second runnerup? Solution: Is the order important? YES!!! Miss Argentina, Miss Belgium, Miss China, Miss Denmark, Miss Egypt 5 X 4 X 3 ___ ___ ___ 1st 2nd 3rd n! nP = r (n − r )!
5! 5! = 5P = 3 (5 − 3)! 2! 5 × 4 × 3 × 2 ×1 = 2 ×1 = 5× 4× 3 = 60 Example 3: In how many ways can a committee of 4 be selected from a group of 15 people? Solution: Is the order important? Solution: Is the order important? NO! It ‘s just who is on the committee So this is a COMBINATION
15 C4 = ()
15 4 15! 15 ×14 ×13 ×12 = = = 1365 4!(11)! 4 × 3 × 2 ×1 Example 4(i): Among 20 applicants for 5 positions advertised on a newspaper, 9 are college graduates. If the selection is random, what is the probability that the positions will be filled by 3 graduates and 2 non graduates? Example 4(ii): An auto parts store has 200 rebuilt starters in stock, of which 4 are HYPERGEOMETRIC HYPERGEOMETRIC DISTRIBUTION DISTRIBUTION
Solution: Example 4(i) ( )( ) = 84 x55 = 0.29799 P( E ) = ( ) 15504
9 3 11 2 20 5 Solution: Example 4(ii) ( )( ) = 6 x196 = 0.0009 P( E ) = ( ) 1,313,400
4 2 196 1 200 3 SUMMARY: SUMMARY: HYPERGEOMETRIC DISTRIBUTION HYPERGEOMETRIC
Let X = the number of x “successes” in “n” trials then X ~ Hypergeom (N, R, n, x) where n = the number of trials N = the number of elements in the population R = the number of elements labeled “success” Note: This is the only distribution we study that is “WITHOUT REPLACEMENT” ( )( ) P( X = x) = ()
R x N −R n− x N n for 0 ≤ x ≤ R ...
View
Full
Document
 Spring '10
 Unknown

Click to edit the document details