Unformatted text preview: 178 Solutions to Exercises 14. [BB] Let box 1 correspond to days 1, 2, and 3, box 2 to days 4,5, and 6, box 3 to days 7, 8, and 9, and box 4 to days 10, 11, and 12. Putting each bill into the box corresponding to the day it was mailed, we see that one box must contain at least 11~51 = 49 bills, by the general form of the PigeonHole Principle. This gives the desired result. 15. First note that 1 + 2 + 3 + ... + 36 = 36~37) = 666. Since there are twelve blocks of three consecutive sectors, there must be a sum of at least 1616261 = 56 in one of these blocks. (There are 666 pigeons shoved into 12 pigeon holes.) 16. By the multiplication rule, there are 26 2 possible pairs of initials. If n seats are occupied, the Pigeon Hole Principle (strong form) says that 1 2~2 1 people seated will have the same first and last initials. The smallest n such that 1 2~2 1 = 3 is n = 2(26 2 ) + 1 = 1353. 17. [BB] Divide the rectangle into 25 rectangles, each 3 x 4. By the PigeonHole Principle, at least two 17....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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