This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: University of California, Los Angeles Department of Statistics Statistics 100A Instructor: Nicolas Christou Combinatorial analysis Basic principle of counting: Suppose two experiments are to be performed. Then, if the first experiment can result in m outcomes and if for each outcome of the first experiment there are n outcomes of the second experiment, then all together there are m × n possible outcomes. Examples: Permutations: How many different ordered arrangements of the letters A,B,C are possible? There are are 6 permutations : ABC,ACB,BAC,BCA,CAB,CBA . Or, using the basic principle of counting we can find the number of permutations as follows: 3 × 2 × 1 = 6. In general n objects can be ordered in n × ( n 1) × ( n 2) × ··· × 1 = n ! ways. Each arrangement it is called a permutation . Example: In how many ways can 4 math books, 3 chemistry books, and 2 history books can be ordered so that books of the subject are together? Suppose k objects are to be selected and ordered from n objects ( k < n ). Say, n = 4 ( A,B,C,D ), and k = 3. Let’s list all the possible permutations: ABC BCD CDA DAB ABD BCA CDB DAC ACB BDA CAB DBC ACD BDC CAD DBA ADB BAC CBD DCA ADC BAD CBA DCB As we observe there are 24 permutations. Much easier, we can find the number of permuta tions using the basic principle of counting as follows: 4 × 3 × 2 = 24. In general, the number of ways that k objects can be selected and ordered from n objects are: n × ( n 1) × ( n 2) × ··· × ( n k + 1). This can be simplified if we multiply and divide by (...
View
Full
Document
This note was uploaded on 01/14/2011 for the course STATS 100A taught by Professor Wu during the Fall '07 term at UCLA.
 Fall '07
 Wu
 Statistics, Counting

Click to edit the document details