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Unformatted text preview: 201001 Math 222 Sample Final Exam Solution Ideas 1. (a) 7! 2!3!2! (b) Let x i be the number of times the digit i appears. Then we want the number of integer solutions to x + x 1 + ... + x 9 = 7 subject to x i ≥ , ≤ i ≤ 9, which equals 7+9 1 7 . 2. The LHS counts the number of ways for two people to be chosen from among n men and n women. The RHS counts the same thing by cases. There are n 2 ways in which both people chosen can be men, and n 2 ways in which they can both be women. There are n 1 2 = n 2 ways in which the chosen people can be of opposite gender. Hence, by the AP, the number of ways the two people can be chosen is the RHS: 2 n 2 + n 2 . Therefore 2 n 2 = 2 n 2 + n 2 . 3. There are 50 odd numbers among 1 , 2 ,..., 100. Partition { 1 , 2 ,..., 100 } into sets by placing two numbers in the same set when they have the same largest odd divisor. Since 51 integers are selected, the Pigeonhole Principle guarantees that some two of them have the same largest odd divisor, t . Let these two numbers be a and b , where a < b . Then, a = 2 r t and b = 2 s t , where r < s , so that a 2 s r = b . Since s r is a positive integer, it follows that a  b . 4. (a) Let x i be the number of rings on finger i, 1 ≤ i ≤ 4. We want the number of solutions to x 1 + x 2 + x 3 + x 4 = 8 , x i ≥ , 1 ≤ i ≤ 4 , which is 8+3 8 . (b) As in part (a) but x i ≥ 1 , 1 ≤ i ≤ 4, so the solution is 4+3 4 . (c) This is the number of functions from rings to fingers, which is 84 8 . (d) This is the number of sequences of 8 rings and three separators, which is 11!...
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This note was uploaded on 01/15/2012 for the course MATH 222 taught by Professor Huangjing during the Spring '11 term at University of Victoria.
 Spring '11
 HuangJing
 Math

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