math3012_hw6
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math3012_hw6

Course Number: GTG 977, Fall 2009

College/University: Georgia Tech

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MATH 3012 Homework 6 Andrew Kerr gtg977g March 8, 2005 Section 8.1 6.) Determine how many integer solutions there are to x1 +x2 +x3 +x4 = 19 if b.) 0 xi < 8 for all 1 i 4 By the principle of inclusion and exclusion, N (c1 c2 c3 c4 ) = S0 - S1 + S2 - S3 + S4 . We let S0 equal 19 + 4 - 1 19 = 1540 and we define the following conditions: c0 : x1 8, c1 : x2 8, c2 : x3 8, c3 : x4 8 S1 S2 S3 S4 = N (c1 ) +...

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3012 MATH Homework 6 Andrew Kerr gtg977g March 8, 2005 Section 8.1 6.) Determine how many integer solutions there are to x1 +x2 +x3 +x4 = 19 if b.) 0 xi < 8 for all 1 i 4 By the principle of inclusion and exclusion, N (c1 c2 c3 c4 ) = S0 - S1 + S2 - S3 + S4 . We let S0 equal 19 + 4 - 1 19 = 1540 and we define the following conditions: c0 : x1 8, c1 : x2 8, c2 : x3 8, c3 : x4 8 S1 S2 S3 S4 = N (c1 ) + N (c2 ) + N (c3 ) + N (c4 ) = N (c1 c2 ) + N (c1 c3 ) + N (c1 c4 ) + N (c2 c3 ) + N (c2 c4 ) + N (c3 c4 ) = N (c1 c2 c3 ) + N (c1 c2 c4 + N (c1 c3 c4 ) + N (c2 c4 c4 ) = N (c1 c2 c3 c4 ) Due to symmetry of the conditions, we find 19 N (c1 ) = N (c2 ) = N (c3 ) = N (c4 ) = i=8 3 + 19 - i - 1 19 - i = 364 and S1 = 4N (c1 ) = 1456. We see that N (c1 c2 ) = N (c1 c3 ) = N (c1 c4 ) = N (c2 c3 ) = N (c2 c4 ) = N (c3 c4 ) = 20 1 and S1 = 6N (c1 c2 ) = 120. We also find that any three conditions cannot be simultaneously true. Therefore, S3 = S4 = 0. N (c1 c2 c3 c4 ) must equal 1540 - 1456 + 120 = 204. c.) 0 x1 5, 0 x2 6, 3 x3 7, and 3 x4 8. We define the following four conditions: c1 : x1 6, c2 : x2 7, c3 : 0 x3 2 or x3 8, c4 : 0 x4 2 or x4 9 As in part (b), N (c1 c2 c3 c4 ) = S0 - S1 + S2 - S3 + S4 where S1 = N (c1 ) + N (c2 ) + N (c3 ) + N (c4 ) S2 = N (c1 c2 ) + N (c1 c3 ) + N (c1 c4 ) + N (c2 c3 ) + N (c2 c4 ) + N (c3 c4 ) S3 = N (c1 c2 c3 ) + N (c1 c2 c4 + N (c1 c3 c4 ) + N (c2 c4 c4 ) S4 = N (c1 c2 c3 c4 ) However, symmetry does not exist among the conditions, and each of the terms must be computed explicitly. Using the rules of sum and product, the following terms may be evaluated: N (c1 ) = 560 N (c4 ) = 857 N (c1 c4 ) = 309 N (c3 c4 ) = 535 N (c1 c3 c4 ) = 149 N (c1 c2 c3 c4 ) = 101. There are 101 integer solutions to x1 + x2 + x3 + x4 = 19 if 0 x1 5, 0 x2 6, 3 x3 7, and 3 x4 8. N (c2 ) = 455 N (c1 c2 ) = 84 N (c2 c3 ) = 270 N (c1 c2 c3 ) = 64 N (c2 c3 c4 ) = 185 N (c3 ) = 935 N (c1 c3 ) = 330 N (c2 c4 ) = 255 N (c1 c2 c4 ) = 64 N (c1 c2 c3 c4 ) = 45 16.) How many social security numbers (nine-digit have sequences) each of the digits 1, 3, and 7 appearing at least once? 2 We define the following conditions: c1 : xi = 1 for 1 i 9, c2 : xj = 3 for 1 j 9, c3 : xk = 7 for 1 k 9 We let N (c1 c2 c3 ) = S0 - S1 + S2 - S3 where S0 = 109 , S1 = N (c1 ) + N (c2 ) + N (c3 ), S2 = N (c1 c2 ) + N (c1 c3 ) + N (c2 c3 ), and S3 = N (c1 c2 c3 ). Because symmetry exists among the conditions, N (c1 ) = N (c2 ) = N (c3 ) = 99 , N (c1 c2 ) = N (c1 c3 ) = N (c2 c3 ) = 89 , and N (c1 c2 c3 ) = 79 . We therefore find N (c1 c2 c3 ) = 109 - 3 99 + 3 89 - 79 = 200038110. 20.) At a 12-week conference in mathematics, Sharon met seven of her friends from college. During the conference, she met each friend at lunch 35 times, every pair of them 16 times, every trio eight times, every foursome four times, each set of five twice, and each set of six once, but never all seven at once. If she had lunch every day during the 84 days of the conference, did she ever have lunch alone? We see that 84 - 35 7 1 + 16 7 2 -8 7 3 +4 7 4 -2 7 5 + 7 6 =0 By the principle of inclusion and exclusion, we see that she did not once dine alone. Section 8.2 2a.) In how many ways can the letters in ARRANGEMENT be arranged so that there are exactly two pairs of consecutive identical letters? at least two pairs of consecutive identical letters? 2b.) Answer part (a), replacing two with three. 3 6.) Zelma is having a luncheon for herself and nine of the women in her tennis league. On the morning of the luncheon she places name cards at the ten places at her table and then leaves to run a last-minute errand. Her husband comes home from his morning tennis match and leaves the back door open. A gust of wind scatters the ten name cards. In how many ways can the ten cards be replaced so that exactly four of the ten women will be seated where Zelma had wanted them? In how many ways will at least four of them be seated where they were supposed to be? Section 8.3 6.) How many derangements of 1,2,3,4,5,6,7,8 start with (a) 1,2,3, and 4 in some order? (b) 5, 6, 7, and 8 in some order? 4
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