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# HW10 - HW10 Solutions Dilip Raghavan 1 Section 8.2 Problem...

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HW10 Solutions Dilip Raghavan November 23, 2008 1 Section 8.2 Problem 12. Let U be the set of all 4–digit numbers, including those with leading 0s. For each i between 0 and 9, let A i be the set of all 4–digit numbers in which the digit i occurs exactly twice. Now, |U| = 10 4 . For each i , | A i | = 4 2 × 9 2 – first choose the 2 slots for the exactly 2 occurances of i , and then arrange any of the other 9 digits in the two remaining slots. Therefore, S 1 = 10 1 × 4 2 × 9 2 . Now, given a pair i j , A i A j = 4 2 – choose 2 slots for the two occurrences of i , and then j must go into the remaining 2 slots. Therefore, S 2 = 10 2 × 4 2 . Notice that all triple intersections are empty because a 4–digit number cannot have 3 digits ocurring exactly twice. Thus, ¯ A 0 ∩ · · · ∩ ¯ A 9 = |U|- S 1 + S 2 = 10 4 - 10 1 × 4 2 × 9 2 + 10 2 × 4 2 . Problem 18. For a real number x , let x denote the least integer greater than or equal to x . Note that the number of odd numbers in the set { 1 , . . . , n } is equal to n 2 . Now, let U be the set of all possible arrangements of 1 , 2 , . . . , n . For each odd number i between 1 and n , let A i be the set of arrangements with i ocurring in slot i (so i is not deranged). |U| = n !. For each odd number i , |

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HW10 - HW10 Solutions Dilip Raghavan 1 Section 8.2 Problem...

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