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Perms_Combs - ECON*2740ClassNotes:...

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1 ECON*2740 Class Notes: Permutations and Combinations Course Instructor: David Prescott Permutations and Combinations A permutation is an ordered set of objects. For example, consider the numbers 1,2 and 3. These three numbers can be arranged in six different orders or permutations: (1) 1 2 3 (2) 1 3 2 (3) 2 1 3 (4) 2 3 1 (5) 3 1 2 (6) 3 2 1 The number of permutations can be calculated as follows: the “1” can be placed in any one of 3 positions. Once that position is selected, the “2” can be placed in either of 2 positions and finally the “3” has to be placed in the one remaining position. The total number of permutations is therefore 3x2x1 = 6 The concept of “permutation” is helpful in calculating many types of probabilities. For example, suppose three tokens labelled 1, 2 and 3 are placed in a random order. What is the probability that “1” is in the first position? The first step is to define the sample space, which is the set of equally likely outcomes. We have seen that in this case, the sample space consists of 6 unique orderings. Each has a probability of 1/6 of occurring. The event A can be defined as: “1” appears in the first position. We can see by examining the sample space that in two cases “1” appears in the first position, so the set (event) A has two elements and: Prob(A) = 2/6 = 1/3 The number of basic outcomes in A can also be calculated. When “1” is placed in the first position, the “2” can appear in either the second or third position and finally the “3” has to occupy the one remaining open slot. Hence, the event A has two elements. Additional Examples: 5 tokens are marked 1 to 5. They are selected randomly one after the other and placed in the order in which they are drawn.
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2 Q1: Find probability that they are selected in the specific order 1, 2, 3, 4, 5 (event A) Q2: Find the probability the first two to be drawn are 5 and 4 in that order (event B) First, define the sample space S – the set of basic outcomes that are all equally probable. This is the number of ways the 5 tokens can be placed in order. The first token can be any one of the 5. Once this has been selected, the second can be chosen in any one of 4 ways, the third in any one of 3 ways and so one. The total number of ordered arrangements of 5 different tokens is 5x4x3x2x1 = 120. This product is usually written as 5! The event that the tokens are drawn in the specific order 1 to 5 represents just 1 of the 120 possible orderings, the probability of event A is 1/120 In general, n different objects can be ordered in n(n 1)(n 2)..(2)(1) = n ! different ways. The term n ! is referred to as n factorial Q2: Find the probability the first two tokens are 5 and 4 in that order (event B) Event B allows the 3 rd , 4 th and 5 th tokens to be in any order provided the 1 st and 2 nd are 5 and 4 respectively. Token 3 can be selected in any of 3 ways, token 4 in any of 2 ways and token 5 will be the one left over. That’s 3x2x1 possible orderings.
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