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00 r 0s and n1 s the total number of ways is r

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Unformatted text preview: and “|” in, say, 00|000||00|0|||00.....|0||0 r "0"s and n−1 "|"’s The total number of ways is r +n−1 r 4 = (r + n − 1)! . r !(n − 1)! Unordered selection without replacement: Consider drawing r balls from n distinct balls without replacement and without regard to order. The total number of ways is n r = n! . r !(n − r )! Review: Combinatorial Analysis (cont’d) III Arrangement of n objects, with r distinct types. In other words, there are n1 objects of type 1, n2 objects of type 2, ..., nr objects of type r . (e.g. number of different letter arrangements can be formed using the letters STATISTICS.) Total number of ways is n! . n1 ! × n2 ! × · · · × nr ! Partition of n distinct objects into r distinct groups with specified size n1 , n2 , · · · , nr . (e.g. number of ways in dividing a class of 40 into groups of 10, 10 and 20.) Toal number of ways is n n1 , n2 , · · · , nr = n! . n1 ! × n2 ! × · · · × nr ! Review: Combinatorial Analysis (cont’d) IV Partition of n indistinguishable objects into r distinct groups (i.e. we only concern the number of objects in each group). The problem is the same as arranging the “0” and “|” in, say, 00|000||00|0|||00.....|0||0 n "0"s and r −1 "|"’s Total number of ways is n+r −1 n = (n + r − 1)! . n!(r − 1)! Problem 1 Problem 1. A product code of manufacturer consists of 2 letters followed by 3 digits. The manufacturer only uses the letters A, B, C and D. For example, AA032 and BD369 are valid product codes. If a product is selected at random, what is the probability that 1 2 3 4 it begins with the letter A? the first two letters are identical? the letters are identical and the digits are equal? the letters are formed by B and D, and the 3-digit number is even? Solution to Problem 1 Solution. 1 2 3 4 Since the four letters are equally likely to be the first letter, P(beginning with A) = 1 . 4 4 P(two letters are identical) = 4×4 = 1 . 4 1 P(the digits are equal) = 10×1...
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