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University of KwaZulu-Natal, Pietermaritzburg Campus School of Mathematical Sciences MATH 236: TEST 1 Solutions Time Allowed: 90 minutes Instructions: Answer all the questions and show all working. Total marks: 50 Total questions: 6 Bonus marks: 5 Question 1. [Total: 10] (1) State without proof the Addition Principle (5 points) (2) Let S and T be finite sets. Using Addition Principle to prove that |S × T | = |S| · |T |. (5 points) Solution (1) If {A1 , A2 , · · · , Ak } is a pairwise disjoint collection of finite sets, then | ∪ki=1 Ai | = k X |Ai |. i=1 (2) Let S = {x1 , x2 , · · · , xs } and T = {y1 , y2 , · · · , yt }, where s = |S| and t = |T |. Let Ai be the set of ordered pairs whose first element is xi , that is, Ai = {(xi , y) : y ∈ T }. Clearly, S × T = ∪si=1 Ai and |Ai | = t for 1 ≤ i ≤ s. Also, {A1 , A2 , · · · , As } is a pairwise disjoint collection of subsets. Thus by Addition Principle, we have s X |S × T | = |A1 ∪ A2 ∪ · · · ∪ As | = |Ai | = st = |S| · |T |. i=1 Question 2. [Total: 10] Let R be a relation defined on the set S = {4, 5, 6, 11, 12, 13, 18} by xRy if and only if 7 | 3x + 4y. (1) Show that R is an equivalence relation. (6 points) (2) Find [4], the equivalence class containing 4. (2 points) (3) Find |S/R| (2 points) Solution. (1) For any x ∈ S, we have that 3x + 4x = 7x is divisible by 7 so xRx, so R is reflexive. Assume that xRy where x, y ∈ S. Then 7 | 3x + 4y. Since 7(x + y) − (3x + 4y) = 4x + 3y and both 7(x + y) and 3x + 4y are divisible by 7, we deduce that 7 | 3y + 4x, so yRx and hence R is symmetric. Finally, assume that xRy and yRz for some x, y, z ∈ S. Then 3x + 4y = 7s and 3y + 4z = 7t for some s, t ∈ Z. It follows that 3x + 4y + 3y + 4z = 7(s + t) and hence 3x + 4z = 7(s + t − y) is divisible by 7. Thus xRz and therefore R is transitive. Since R is reflexive, symmetric and transitive, it is an equivalence relation. (2) We have [4] = {y ∈ S : 4Ry} = {y ∈ S : 7 | 3 · 4 + 4y} = {4, 11, 18}. 2 (3) We have that [5] = {5, 12}, [6] = {6, 13} and S = [4] ∪ [5] ∪ [6]. So |S/R| = 3. Question 3. [Total: 10] (1) How many ways can the letters of the word ECCEDENTESI be arranged in a letters TESI must remain together as a block? (2) How many ways can the committee of 6 be chosen from a group of 15 people if the consists one president, two secretaries and 3 members? row if the (5 points) committee (5 points) Solution. (1) Consider ‘TESI’ as one block, there are 8 total letters, in which there are 3 letters ‘E’, 2 letters ‘C’, one letter ‘D’, on e letter ‘N’ and one letter ‘TESI’. Using the formula for permutations of indistinguishable objects, there are exactly 8! = 3360 P (8; 3, 2, 1, 1, 1) = 3! · 2! · 1! · 1! · 1! ways. (2) There are (15)1 ways to choose a president from a group of 15 people; among 15 − 1 = 14 people left, there are 14 2 ways to choose 2 secretaries and finally among 14 − 2 = 12 people, there are 12 3 ways to choose 3 committee members. Thus by the multiplication principle, there are 14 12 (15)1 · · = 300300 2 3 ways. Question 4. [Total: 10] (1) How many odd integers are there from 100 to 999 ? (5 points) (2) How many solutions does the equation x1 + x2 + x3 = 9 have, where x1 , x2 , x3 are nonnegative integers? (5 points) Solution. (1) Every odd number from 100 to 999 has the form abc where a, b, c ∈ {0, 1, · · · , 9} and a 6= 0. Since abc is odd, there are only 5 possibilities for c, since a 6= 0, there are only 9 possibilities for a and finally, there are 10 possibilities for b. Thus by the multiplication principle, there are 9 · 10 · 5 = 450 odd numbers from 100 to 999. (2) The numbers of the nonnegative solutions to the equation x1 + x2 + x3 = 9 is the number of 9-combinations from a set of size 3. Thus there are 3+9−1 11 11 = = = 55 9 9 2 solutions. Question 5. [Total: 10] (1) Find the coefficient of x6 y 5 in the expansion of (2x2 − 3y)8 (5 points) (2) How many numbers between 1 and 84 inclusive, are divisible by at least one of the prime numbers 2, 3 and 7? (5 points) Solution. 3 (1) Let X = 2x2 and Y = −3y. Then (2x2 3y)8 = (X + Y )8 . By Binomial Theorem, the coefficient of X 3 Y 5 in the expansion of (X + Y )8 is 83 and since X 3 Y 5 = (2x2 )3 (−3y)5 = 23 (−3)5 x6 y 5 , we deduce that the coefficient of x6 y 5 in the expansion of (2x2 − 3y)8 is 8 23 · (−3)5 · = −108864. 3 (2) Let S = {1, 2, · · · , 84}. Let S2 be a subset of S consisting of all numbers divisible by 2, S3 be a subset of S consisting of all numbers divisible by 3, and finally S7 be a subset of S consisting of all numbers divisible by 7. We have |S2 | = |{x ∈ S : 2 | x}| = |{x ∈ S : x = 2k, 1 ≤ 2k ≤ 84}| = |{x ∈ S : x = 2k, 1 ≤ k ≤ 84 2 }| = |{x ∈ S : x = 2k, 1 ≤ k ≤ 42}| = 42. Similarly, we have |S3 | = 84 3 = 28 |S7 | = 84 7 = 12 |S2 ∩ S3 | = 84 6 = 14 |S2 ∩ S7 | = 84 14 = 6 |S3 ∩ S7 | = 84 21 = 4 |S2 ∩ S3 ∩ S7 | = 84 42 = 2. By the Inclusion-Exclusion Principle, we have |S2 ∪ S3 ∪ S7 | = |S2 | + |S3 | + |S7 | − |S2 ∩ S3 | − |S2 ∩ S7 | − |S3 ∩ S7 | + |S2 ∩ S3 ∩ S7 | = 42 + 28 + 12 − 14 − 6 − 4 + 2 = 60. Hence, there are exactly 60 numbers between 1 and 84 inclusive which are divisible by at least one of the numbers 2, 3 and 7. Extra Marks. Let n and k be integers with 1 ≤ k ≤ n. Prove the identity n n−1 k =n k k−1 by using a combinatorial proof. (5 points) Solution. Let S be a set consisting of n people. We want to count the number of ways we can choose group of k-people and then choose a leader from this group. Clearly, there are nk groups of k-people from a set of size n. Among each group of k-people, there are k1 = k ways to choose a leader from this group. Therefore, by the Multiplication Principle, there are nk k1 = k nk ways. We next count this number in a different way. We now choose a leader from the whole group of n-people and then among n − 1 people left, we choose a group of k − 1 people withthe chosen leader. Obviously, there are n1 = n ways to choose a leader first and then there are n−1 ways to choose a k−1 n−1 group of (k − 1)- people. Thus by the Multiplication Principle, there are n1 n−1 = n k−1 k−1 ways. Now by the counting principle, we obtain that n n−1 k =n . k k−1