Solutions_4_W

# Solutions_4_W - MAT 312/AMS 351 Summer 2011 Solutions to...

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MAT 312/AMS 351, Summer 2011 Solutions to Homework Assignment 4 Maximal grade for HW4: 120 points Section 4.1 Let π 1 , π 2 , π 3 , π 4 , π 5 be the following permutations: π 1 = 1 2 3 4 5 6 7 8 9 3 2 1 6 5 4 9 8 7 , π 2 = 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 , π 3 = 1 2 3 4 5 6 7 8 9 4 5 6 7 8 9 1 2 3 , π 4 = 1 2 3 4 5 6 7 8 9 10 11 12 9 1 2 3 4 5 6 7 8 11 12 10 , π 5 = 1 2 3 4 5 6 7 8 9 10 11 12 12 7 2 8 4 6 3 9 5 1 11 10 . 1. (32 points) Calculate the following products: π 1 π 2 , π 2 π 3 , π 3 π 1 , π 3 π 2 , π 2 π 1 π 3 , π 2 π 2 π 2 , π 4 π 5 , π 5 π 4 , π 1 π 3 , π 2 π 2 , π 2 π 1 , π 3 π 3 , π 2 π 1 π 2 , π 2 π 3 π 2 , π 4 π 4 , π 5 π 5 Answer: π 1 π 2 = 1 2 3 4 5 6 7 8 9 7 8 9 4 5 6 1 2 3 , π 2 π 3 = 1 2 3 4 5 6 7 8 9 6 5 4 3 2 1 9 8 7 , π 3 π 1 = 1 2 3 4 5 6 7 8 9 6 5 4 9 8 7 3 2 1 , π 3 π 2 = 1 2 3 4 5 6 7 8 9 3 2 1 9 8 7 6 5 4 , 1

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π 2 π 1 π 3 = 1 2 3 4 5 6 7 8 9 4 5 6 1 2 3 7 8 9 , π 2 π 2 π 2 = 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 , π 4 π 5 = 1 2 3 4 5 6 7 8 9 10 11 12 10 6 1 7 3 5 2 8 4 9 12 11 , π 5 π 4 = 1 2 3 4 5 6 7 8 9 10 11 12 5 12
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