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Combinatorics Homework A Name: Sam Poole January 16, 2007 A7 Find the number of sequences of length 4 in which the first two terms are letters from the alphabet {A, B, C, …, Z} and the last two terms are digits from the set {0, 1, 2, …, 9} with the condition that the two digits must be different. Solution. We can imagine filling in four blanks ___ ___ ___ ___, left to right. here are 26 choices for the first blank, 26 choices for the second blank, 10 choices for the third blank and 9 choices for the fourth blank. By the Product Rule, there are (26)(26)(10)(9) such sequences. A19 Suppose the numbers 1, 2, 3, 4, 5 are arranged in a random order. What is the probability that all odd numbers still appear in odd numbered positions. Solution. We can imagine filling in five blanks ___ ___ ___ ___ ___, left to right, to create the (equally likely) outcomes. The total number of these is 5! = 120, by the Product Rule. To count the number of these satisfying the given condition, simply adapt the Product Rule accordingly: Filling in left-to-right, there are 3 choices for position one, 2 choices for position two, 2 choices for position three, 1 choice for position four, and then 1 choice for position five. Hence, the number of outcomes satisfying the given criterion is (3)(2)(2)(1)(1) = 12, which makes the 12 1 probability of this happening . 120 10 A20c Find all derangments of ABCD. Solution. There aren’t that many, so we will just list all 9 of them: BADC DABC BCDA DCAB BDAC DCBA CADB CDBA CDAB A21 Find the number of rearrangements of ABCDE having exactly one letter in its original position. Solution. Suppose the letter A is the letter in its original position. Hence, the arrangement looks like A____ The number of ways to fill in the remaining blanks so that no other letter is in its original position is 9 (the value of D4) according to the solution to problem A20 above. Similarly, if the letter that remains in its original position is B, then there are 9 ways to complete the other four blanks. This is true for any of the five letters that remain in their original position. Therefore, the total number of arrangements is 5 (the number of choices for which letter stays fixed) times 9 (the number of derangements of the other four letters). This gives a final answer of 45.