Feb1_2011 - ECON*2740ClassNotes,Winter2011 Tuesday,1February

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1 ECON*2740 Class Notes, Winter 2011 Tuesday, 1 February Course Instructor: David Prescott Permutations and Combinations These concepts are helpful in calculating many types of probabilities. For example, 5 tokens are marked 1 to 5. They are selected randomly one after the other. 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 – the set of basic outcomes that are all equally probable. For Q1, determine how many 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 Example, consider how many ways 3 tokens numbered 1, 2, 3 can be arranged. The answer is 3! = 3x2x1 = 6: (1) 1 2 3 (2) 1 3 2 (3) 2 1 3 (4) 2 3 1 (5) 3 1 2 (6) 3 2 1
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2 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. Hence the sequence 5, 4, *, *, * (where * means “any token”) can occur in 3! ways out of the 120 possible orderings. Hence P(B) = 6/120 = 1/20 = 0.05 Permutations: The number of ways m different objects can be ordered when selected randomly from n > m objects is called the permutations of m objects drawn from n . The first of the m ordered objects can be any one of the original n objects and the second can be any one of ( n 1 ) and so on. The m th can be selected in any of ( n m+1 ) ways. Hence the number of permutations of m objects drawn from n is : ݊ሺ݊െ1ሻሺ݊െ2ሻ … ሺ݊െ݉൅1ሻ This can also be written as: ܲ ݊ሺ݊െ1ሻ. . ሺ݊െ݉൅1ሻሺ݊െ݉ሻሺ݊െ݉െ1ሻ…1 ሺ݊െ݉ሻሺ݊െ݉െ1ሻ….1 ݊!
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Feb1_2011 - ECON*2740ClassNotes,Winter2011 Tuesday,1February

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