Chapt 1d - 1.10 Basic Combinatorial Analysis 31 1.10 Basic...

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Unformatted text preview: 1.10 Basic Combinatorial Analysis 31 1.10 Basic Combinatorial Analysis Combinatorial analysis deals with counting the number of different ways in which an event of interest can occur. Two basic aspects of combinatorial analysis that are used in probability theory are permutation and combination. 1.10.1 Permutations Sometimes we are interested in how the outcomes of an experiment can be arranged. For example, if the possible outcomes are A, B, and C, we can think of six possible arrangements of these outcomes: ABC, ACB, BAC, BCA, CAB, and CBA. Each of these arrangements is called a permutation . Thus, there are six permutations of a set of three distinct objects. This number can be derived as follows: There are three ways of choosing the first object; after the first object has been chosen, there are two ways of choosing the second object; and after the first two objects have been chosen, there is one way to choose the third object. This means that there are 3 2 1 = 6 permutations. For a system of n distinct objects we can apply a similar reasoning to obtain the following number of permutations: n ( n- 1 ) ( n- 2 ) 3 2 1 = n ! where n ! is read as n factorial. By convention, 0 ! = 1. Assume that we want to arrange r of the n objects at a time. The problem now becomes that of finding how many possible sequences of r objects we can get from n objects, where r n . This number is denoted by P ( n , r ) and defined as follows: P ( n , r ) = n ! ( n- r ) ! = n ( n- 1 ) ( n- 2 ) ( n- r + 1 ) r = 1 , 2 , . . . , n The number P ( n , r ) represents the number of permutations (or sequences) of r objects taken from n objects when the arrangement of the objects within a given sequence is important. Note that when r = n , we obtain P ( n , n ) = n ( n- 1 ) ( n- 2 ) 3 2 1 = n ! Example 1.18 A little girl has six building blocks and is required to select four of them at a time to build a model. If the order of the blocks in each model is important, how many models can she build? 32 Chapter 1 Basic Probability Concepts Solution Since the order of objects is important, this is a permutation problem. Therefore, the number of models is given by P ( 6 , 4 ) = 6 ! ( 6- 4 ) ! = 6 ! 2 ! = 6 5 4 3 2 1 2 1 = 360 trianglesolid Note that if the little girl were to select three blocks at a time, the number of permutations decreases to 120. Example 1.19 How many words can be formed from the word SAMPLE? As- sume that a formed word does not have to be an actual English word, but it may contain at most as many instances of a letter as there are in the original word (for example, maa is not acceptable, since a does not appear twice in SAMPLE, but mas is allowed)....
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This note was uploaded on 01/05/2010 for the course STAT 350 taught by Professor Carlton during the Fall '07 term at Cal Poly.

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Chapt 1d - 1.10 Basic Combinatorial Analysis 31 1.10 Basic...

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