By Marcel B. Finan
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This path was once learn in Brock college via Jan Vrbik.
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Induction Conclusion We must show n(S1 × S2 × · · · × Sk+1 ) = n(S1 ) · n(S2 ) · · · n(Sk+1 ). To see this, note that there is a one-to-one correspondence between the sets S1 ×S2 ×· · ·×Sk+1 and (S1 ×S2 ×· · · Sk )×Sk+1 given by f (s1 , s2 , · · · , sk , sk+1 ) = 3 THE FUNDAMENTAL PRINCIPLE OF COUNTING 33 ((s1 , s2 , · · · , sk ), sk+1 ). 4). 2 In designing a study of the effectiveness of migraine medicines, 3 factors were considered: (i) (ii) (iii) Medicine (A,B,C,D, Placebo) Dosage Level (Low, Medium, High) Dosage Frequency (1,2,3,4 times/day) In how many possible ways can a migraine patient be given medicine?
7 An ice cream store sells 21 flavors. Fred goes to the store and buys 5 quarts of ice cream. How many choices does he have? Solution. Fred wants to choose 5 things from 21 things, where order doesn’t matter and repetition is allowed. 3 gives the number of vectors (n1 , n2 , · · · , nk ), where ni is a nonnegative integer, such that n1 + n2 + · · · + nk = n where ni is the number of objects in box i, 1 ≤ i ≤ k. 8 How many solutions to x1 + x2 + x3 = 11 are there with xi nonnegative integers? Solution.
Objects seated in a row) depending on where we start. Since there are exactly n! linear permutations, there are exactly n! = (n − 1)! circular permutations. However, if the two above permutations n are counted as the same, then the total number of circular permutations of . this type is (n−1)! 6 In how many ways can you seat 6 persons at a circular dinner table. Solution. There are (6 - 1)! = 5! = 120 ways to seat 6 persons at a circular dinner table. Combinations In a permutation the order of the set of objects or people is taken into account.
A Probability Course for the Actuaries: A Preparation for Exam P 1 by Marcel B. Finan