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Unformatted text preview: The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 1 (Week 2) MATH2069/2969: Discrete Mathematics and Graph Theory Semester 1, 2010 1. If there are 40 students in the class, their surnames can’t all start with different letters (because there are only 26 letters). What can be said if there are 80 students in the class? Solution: The Pigeonhole Principle says that there must be d 80 26 e = 4 students whose surnames begin with the same letter. This doesn’t mean that there is nec essarily a letter which ‘has’ exactly 4 students, but there must be a letter which has at least 4. Recall the reasoning: if each of the 26 letters were to have at most 3 students, then there could be at most 26 × 3 = 78 students. 2. How many sixdigit numbers are there (not starting with 0)? How many of these have six different digits? Explain your answers. Solution: The sixdigit numbers are all the numbers greater than or equal to 100000 and less than 1000000, so there are 900000 of them. Another way to count them is to note that there are 9 possibilities for the first digit, and then choosing the other digits is an ordered selection of 5 from 10 possibilities with repetition allowed, so the total is 9 × 10 5 . The number of sixdigit numbers which have no repeated digits is 9 × 9 (5) = 9 × 9 × 8 × 7 × 6 × 5 = 136080, because there are 9 choices for the first digit, and then choosing the other digits is an ordered selection of 5 from 9 possibilities with repetition not allowed. 3. Suppose you have 7 different ornaments to put on your mantelpiece. (a) If you want to use all of them, how many possible arrangements are there? Solution: There are 7 choices for which ornament goes first, then 6 choices for which goes next, then 5, etc. So the answer is 7! = 5040. (b) If you want to use 6 of them, how many possible arrangements are there? Solution: By the same reasoning as in the previous part, the answer is 7 × 6 × 5 × 4 × 3 × 2, which is also 5040. It makes sense that the answer should be the same as the previous part, because you can imagine storing the unused ornament at the righthand edge of the mantelpiece and thus ‘using’ it after all. (c) If you can use all, some, or none of them, how many possible arrangements are there? Solution: If you use k of the ornaments, then by the same reasoning as in the previous parts, the number of possible arrangements is 7 ( k ) = 7! (7 − k )! . So the answer is 7! 0! + 7! 1! + 7! 2! + 7! 3! + 7! 4! + 7! 5! + 7! 6! + 7! 7! = 13700 . Copyright c ° 2010 The University of Sydney 1 *(d) Divide your answer to part (c) by your answer to part (a); this ratio measures how much extra freedom you get by not necessarily using all the ornaments....
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 Spring '09
 Math, Statistics, Graph Theory, Natural number, ﬁrst digit

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