# Spring 2007 Math 510 HW5 Solutions

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1 Spring 00 Math 0 HW Solutions Section 6.. Find the number of integers between and 0,000 inclusive which are not divisible by, 6,, or 0. Solution. Let S = {,,..., 0000}. Set A = the set of numbers in S that are divisible by ; A = the set of numbers in S that are divisible by 6; A = the set of numbers in S that are divisible by ; A = the set of numbers in S that are divisible by 0. The numbers we are counting are exactly those numbers that are in S but not in any one of A, A, A, and A. We will use the inclusion-exclusion principle. Note that S = 0000 and A = {n n 0000}. Thus A = 0000 = 00, A = 0000 = 666, 6 A = 0000 = 8, A = = 000. Note that A A consists of those numbers in S that are both multiples of and 6. Thus those numbers that multiples of the least common multiples of and 6 i.e.,. Hence A A = 0000 = 8, A A = =, A A = = 00, A A = 0000 = 8, A A = =, A A = =, A A A = = 9, A A A = = 66, A A A = =, A A A = =, A A A A = =. Thus that total number is = 9.. Determine the number of -combinations of the multiset T = { a, b, c, d}. Solution. We use inclusion-exclusion principle. Set T = { a, b, c, d}. Let A be the set of all -combinations of T with at least a; A be the set of all -combinations of T with at least b;

2 A be the set of all -combinations of T with at least c; A be the set of all -combinations of T with at least 6 d. Let S be the set of all -combinations of T. Then the -combinations of the multiset T is exactly those -combinations of T that has at most a, b, c and d, i.e., the set Ā Ā Ā Ā. + We will use the inclusion-exclusion principle. Note that S = = and the -combinations in A can be achieved by first choosing a and the rest of the objects can be arbitrarily taken from the four types. Thus A = = 0, A = = 6, A = = 0, A = = 8. 6 Note that the -combinations in A A can be obtained by first taking a and b. Then choose the remaining objects from the four types without any restrictions. Thus + + A A = = 0, A A = = 0, + + A A = =, A A = = 0, + + A A = = 0, A A = =. Any -combinations in A A A have to have at least a, b, and c. That is impossible. So we have A A A = 0, A A A = 0, A A A = 0, A A A = 0. Finally, A A A A = 0 and the number of -combinations of T is =. 6. A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 6 chocolate, 6 cinnamon, and plain. If a box contains doughnuts, how many different boxes of doughnuts are possible? Solution. We have the multiset T = {6 a, 6 b, c} here a, b, and c are chocolate, cinnamon, and plain doughnuts respectively. Then the number of different boxes of is exactly the number of -combinations of the multiset T. Set T = { a, b, c}. Let S be the set of all + -combinations of the T. Then S = =. Set A = the set of all -combinations in S with at least a, A = the set of all -combinations in S with at least b, A = the set of all -combinations in S with at least c.

3 Then Ā Ā Ā is exactly the set of all -combinations of T. Note that + + A = =, A = =, 8 + A = =, A 8 A = 0, + + A A = =, A A = =. Also one sees that A A A consists those boxes with with at least a, b, and c. That is impossible. Not by inclusion exclusion principle, we have Thus there 0 different boxes. Ā Ā Ā = = Determine the number of solutions of the equation x + x + x + x = 0 which satisfy x 6, 0 x, x 8, x 6. Solution. equation and Set y = x, y = x, y = x, y = x. Then y, y, y, y satisfies the y + y + y + y = 0 = 0 y, 0 y, 0 y, 0 y. Let S be the set of all solutions of the equation y + y + y + y = which satisfy 0 y, 0 + y, 0 y, 0 y. Then S = = 60. Let A be the set of all solutions in S with y 6, A be the set of all solutions in S with y 8, A be the set of all solutions in S with y, A be the set of all solutions in S with y. Then + A = = 0, A = = 6, A = = 6, A = = 6, A A = = 0, A A = = 0, A A = = 0, A A = =, A A = =, A A = = 0.

4 Note that the intersection of any three or four sets A, A, A and A will have not elements. Thus by inclusion-exclusion principle, Ā Ā Ā = = 96. Thus the equation x + x + x + x = 0 has 96 solutions satisfying which satisfy x 6, 0 x, x 8, x 6.. Determine that number of permutations of {,,..., 8} in which no even integers are in their national positions. Solution. Let S be the set of all permutations of {,,..., 8}. Let A be the set of all permutations in S which has in it natural position; A be the set of all permutations in S which has in it natural position; A be the set of all permutations in S which has 6 in it natural position and A be the set of all permutations in S which has 8 in it natural position. Then A = A = A = A = 8!, A A = A A = A A = A A = A A = A A = 8!, A A A = A A A = A A A = A A A = 8!, A A A A = 8!. Then the set of permutations of {,,..., 8} in which no even integers are in their national positions is Ā Ā Ā Ā and the number of the elements is 8!! + 6 6!! +! = 0.. Determine the number of permutations of {,,..., 9} in which at least one odd integer is in its natural position. Solution. Let S be the set of all permutations of the set {,,..., 9}. Set A = the set of permutations in S such that is in its natural position; A = the set of permutations in S such that is in its natural position; A = the set of permutations in S such that is in its natural position; A = the set of permutations in S such that is in its natural position; A = the set of permutations in S such that 9 is in its natural position. Then the set of permutations of with at least one odd number in its natural position is exactly A A A A A. Note that A = A = A = A = A = 9!, A i A j = 9!, with distinct i, j, A i A j A k = 9! with distinct i, j, k, A i A j A k A l = 9! with distinct i, j, k, l, A i A j A k A l A n = 9! with distinct i, j, k, l, n. Then by the Corollary to the inclusion exclusion principle, A A A A A = 8!! + = 8. 6!! +!

5 Another way to solve this problem to use the logic: the contrary to the fact that at least one odd number is in its natural position is that no odd number is in its natural position. Thus the answer can obtained by subtracting the number of permutations with no odd number in its natural position, which can be computed in a similar way as in Prob. from the total number of all permutations 9!.. At a party, gentlemen check their hats. In how many ways can their hats be returned so that a No gentleman receives his own hat? b at least one of the gentleman receives his own hat? c at least two gentleman receives their own hats? Solution. a. This is D = 8. b.! D =! 8 = 86. c. Let A i be the set of all permutations of the set {,,..., n} having exactly i numbers in their own natural positions. Then A i = n i D n i. Then A 0 A is the set of all permutations n D n. with at most number in its natural position. Then A 0 A = A 0 + A = D n + Thus the answer to c is! D D 6 =.

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