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slides_1_probability

# 87 there are four different scenarios for counting

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87

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There are four different scenarios for counting: Without Replacement With Replacement Ordered Unordered We will fill in these boxes assuming that we have n objects and we would like to select k n of them. 88
Often pays to break up counting into pieces. Think of a “job” consisting of k “tasks.” The fundamental theorem of counting tells us the total number of ways a job can be done. If each task j can be done in n j ways then the entire job can be done in n 1 n 2 n k different ways. 89

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EXAMPLE : Consider the license plate example. The “job” is obtaining a complete license plate number and a “task” is choosing a letter or digit. So there are six tasks. The total number of license plates that can be created is 26 26 26 10 10 10 26 3 10 3 17,576,000 (Can use these arguments to estimate the number of registered vehicles in a state. If the state allows digits first followed by letters, it doubles the number of allowable plates.) 90
Selection without replacements leads to the idea of a permutation . Suppose we have n objects we can select from. A list of these objects of length k , selected without replacement, is a permutation of length k . Because we can select among n objects for the first slot, n 1 for the second slot, up to n k 1 for the k th slot, the number of permutations of length k selected from n objects is n n 1  n 2  n k 2  n k 1 . This is sometimes denoted n P k . 91

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A simpler way to write n P k relies on factorials. For any positive integer n , “ n factorial,” denoted n !, is defined as n ! n n 1  2 1. So, for example, 5! 5 4 3 2 1 120. By convention, 0! 1. It is easily seen that we can write n P k n ! n k ! for any integers n 1 and 1 k n . Note that n ! is simply the number of ways to permute n objects. 92
EXAMPLE : (i) Suppose that a basketball locker room has 12 lockers for its 12 players. How many different ways can the lockers be assigned? Because a student can only be assigned one locker, the answer is 12!, which is a very large number: 479,000,000. (ii) Now suppose there are only 9 players on the team. How many different locker assignments are there? This is 12 P 9 : the first player can be assigned to one of 12 lockers, the second player to one of the remaining 11, and so on. 12 P 9 12 11 10 5 4 79,833,600. 93

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