First Midterm Examination

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1 Spring Semester First Midterm Examination 1) How many numbers in the set {1000, 1001, 1002,..., 9999} have sum of digits equal to 17? 2) Let n Z +. Find the negation, converse and contrapositive of the following propositions: ( ) ( ) a) P 1 : If 4 n and 5 n then 40 n b) P 2 : If (12 ) n then (16 ) n and 3 n 3) Show that (2n 1) 2 = n(2n 1)(2n + 1) 3 4) You have three containers. The two smaller ones have volumes of 189 and 119 cm 3. The third one is much larger. You have unlimited amount of water. Can you put exactly a) 6 cm 3 b) 7 cm 3 of water to the large container? How? 5) Consider the set of all 5 letter words, not necessarily meaningful. Define two relations Q and R on that set as follows: (w 1, w 2 ) Q if exactly three letters of w 1 and w 2 are identical (w 1, w 2 ) R if three or more letters of w 1 and w 2 are identical Here, position of the letter is also important. For example: (TRWQA,ARWSA) Q, (BSTUU,BSTUP) / Q (APNBT,AQWYZ) / Q (TRWQA,ARWSA) R, (BSTUU,BSTUP) R (APNBT,AQWYZ) / R a) Is Q an equivalence relation? b) Is R an equivalence relation? Explain your reasons.

2 Answers 1) Distribute 17 balls to 4 boxes such that the first box has at least 1 ball, and any box has at most 9 balls. Give 1 to first, distribute remaining 16 randomly in ( ) = different ways. Now, to find the cases where one box gets too many balls, give 9 to first, or 10 to any other and distribute remaining ones randomly in ( ) ( )( ) = different ways. So the answer is: = 597 ( ) 2) a) Negation: n such that 4 n and 5 n and (40 ) n ( ) ( ) Converse: If 40 n then 4 n and 5 n Contrapositive: If (40 ) n then (4 ) n or 5 n b) Negation: n such that (12 ) ( ) n and 16 n or 3 n Converse: If (16 n and 3 n) then (12 ) n ( ( ) Contrapositive: If 16 n or 3 n) then 12 n 3) If n = 1, we have 1 = 1. Assume it is correct for n = k, in other words: (2k 1) 2 = Add (2k + 1) 2 to both sides to obtain k(2k 1)(2k + 1) (2k 1) 2 + (2k + 1) 2 = [ ] k(2k 1) = (2k + 1) + 2k [ ] 2k 2 + 5k + 3 = (2k + 1) 3 k(2k 1)(2k + 1) 3 + (2k + 1) 2 (k + 1)(2k + 1)(2k + 3) = 3 So it is correct for n = k + 1, therefore it is correct for all positive integers by mathematical induction.

3 4) First, we have to find gcd(189, 119) using Euclid s algorithm: = = = = = 0 We can see that gcd(189, 119) = 7. Therefore it is impossible to obtain 6 (or any number less than 7) using 189 and 119. Now reversing this process we obtain: 7 = = 49 2 (70 49) = = (119 70) = = ( ) = In other words, we can obtain 7 cm 3 water by adding water 8 times with 119 and removing 5 times with ) These relations are not transitive, therefore they are not equivalence relations. For example: (ABCDE,ABCZW) Q and (ABCZW,XYCZW) Q But: (ABCDE,XYCZW) / Q Exactly the same example also proves R is not transitive.

4 Spring Semester Second Midterm Examination 1) A factory produces a certain model of a car with the following options: 8 different colors. 3 different engines. with or without sunroof. 4 different music systems. Other properties of these cars are identical. This month, the factory produced 400 cars. Which of the following statements (about this month s production) is always correct? Explain. All cars are different. At least 2 cars are identical. At least 3 cars are identical. At least 4 cars are identical. 2) 7 students will sit at a round table. 3 are from Ankara, 2 are from İstanbul and 2 are from İzmir. To meet new people, nobody wants to sit next to someone from the same city with himself/herself. In how many different ways can these students be seated? 3) There are 8 distinct modules to be written for a large software project. There are 10 distinct programmers available. You have to assign the programmers to modules such that: There will be at least one programmer working on each module. More than one programmers may work on a module. A programmer can work on a single module only. Each programmer must work on something. In how many different ways can we do this assignment?

5 4) In how many different ways can we distribute 100 identical balls to 4 different kids such that each kid gets at least 3 and at most 30 balls? 5) We will form an 8-letter string using the letters x, y, z, u, w with the following conditions: There must be at least one x. There must be at least two y s. Number of u s must be even. Number of w s must be odd. How many different strings are possible?

6 Answers 1) = = 2.08 = At least 3 cars are identical. At least 2 cars are identical. 2) Let s call these students a 1, a 2, a 3, i 1, i 2, z 1, z 2. The unwanted couples are: 1. a 1 a 2 2. a 1 a 3 3. a 2 a 3 4. i 1 i 2 5. z 1 z 2 Using inclusion-exclusion, we obtain: 6! 5 5! 2 (All permutations - unwanted couples) + 7 4! ! 2 (+2 unwanted couples) 3 3! ! 2 2 ( 3 unwanted couples) + 3 2! 2 3 (+4 unwanted couples) = 96 Second Method: Place a 1. Choose one arrangement: a 1 a 2 a 3 or a 1 a 3 a 2. (2 options). There will be 3 blanks after a 1. These can be or or ( 3 options) You have to place one i and one z to the double blank. Choose one from each set. (4 options) Order your choice. ( 2) Then order the remaining one-blank elements ( 2 4 options) There is a total number of = 96 options.

7 3) This is equivalent to the number of onto functions from a 10-element set to an 8-element set: 8 10 ( ) ( ) ( ) ( ) ( ) ( ) ( ) = Second Method: Using Stirling table, we find that 8 groups from 10 elements can be made in: = 750 different ways. But here, groups are distinct, so the answer is: 750 8! = ) (x 3 + x x 30 ) 4 = x 12 (1 + x + + x 27 ) 4 ( ) 1 x = x x In this expansion, the coefficient of x 100 is: ( ) ( ) ( ) ( ) = = x 12 ( 1 4x x 56 4x 84 + x 112 ) n=0 ( n + 3 n ) x n

8 5) The series expansion representing the solution is: ) ( (x + x2 x 2 2! + 2! + x3 3! + ) (1 + x + x2 2! + ( ) ( ) e = (e x 1) (e x 1 x) e x x + e x e x e x 2 2 = 1 ) (e 5x 2e 4x + e 3x e x + 2 e x x ) (e 4x e 3x 1 + e x 4 4 ) ) ) (1 + (x x2 2! + + x3 3! + The coefficient of x8 8! is: 1 ( ) ( ) =

9 Second Solution: To see how much series expansion simplify this problem, we can actually count all cases. If we choose, for example, 0u, 1v, 1x, 2y and 4z, there will be 8! 0! 1! 1! 2! 4! = 840 different strings made from these elements. u v x y z strings

10 Spring Semester Final Examination 1) Show that, for n 1: 1 (1!) + 2 (2!) + + n (n!) = (n + 1)! 1 2) There are 10 students in a class: Alperen, Bihter, Cem, Damla, Ege, Gözde, Merve, Saygın, Tolga and Utku. We will select a project team of 4 students, but: Bihter and Damla do NOT want to be in the same team. Damla and Merve do NOT want to be in the same team. Merve and Bihter do NOT want to be in the same team. Tolga and Utku want to be in the same team. In how many different ways can we choose the team satisfying all these constraints? 3) Consider strings of length n generated using the alphabet {a, b, c, d, e, f, g, h} with the following constraints: Number of a s is odd. Number of b s and c s are even. There is at least one d. There are at least two e s. Determine the exponential generating function giving the number of such strings. (You don t have to find the coefficient of xn, just find the function) n! 4) You have to process n data items. You have two available options: Algorithm A finishes the job in 0.5 n 2 operations. Algorithm B finishes the job in 200 n 3/2 operations. For which values of n would you prefer A and for which values of n would you prefer B? (Assuming you want to finish as quickly as possible) 5) Find the solution of the recurrence relation: 6a n a n+1 + 4a n = 0, a 1 = 16, a 2 = 18

11 Answers 1) For n = 1, we have 1(1!) = 2! 1 So the claim is correct for n = 1. Assume it is correct for n = k. 1 (1!) + 2 (2!) + + k (k!) = (k + 1)! 1 Add (k + 1)(k + 1)! to both sides to obtain: 1 (1!) + 2 (2!) + + k (k!) + (k + 1) (k + 1)! = (k + 1) (k + 1)! + (k + 1)! 1 = (k + 2) (k + 1)! 1 = (k + 2)! 1 It is also correct for n = k + 1. So, the claim is correct for all n 1 by mathematical induction. 2) Case 1 - Tolga and Utku are IN: ( ) 8 Choose 2 students from the remaining 8 students in ways. Subtract the cases where you 2 choose Bihter - Damla, or Damla - Merve, or Bihter - Merve: ( ) 8 3 = 25 2 Case 2 - Tolga and Utku are OUT: ( ) 8 Choose 4 students from the remaining 8 students in ways. Subtract the cases where you 4 choose Bihter - Damla and two others (except Merve) etc., and Bihter - Damla - Merve and one other student: ( ) ( ) ( ) = So the answer is: = 60

12 3) ( ) ( ) e x e x e x + e x 2 (e x 1) (e x 1 x) e 3x 2 2 4) They are equally good for 0.5 n 2 = 200 n 3/2 n 1/2 = 400 n = Considering 200 n 3/2 = O ( 0.5 n 2), we should choose A for n < and B for n > ) Assuming a n = x n 6x x + 4 = 0 x = 11 ± x = 11 ± 5 12 x = 1 2 OR x = 4 3 ( a n = c 1 1 ) n ( + c 2 4 ) n 2 3 n = 1 16 = c 1 2 4c 2 3 n = 2 18 = c c 2 9 c 1 = 8, c 2 = 9 ( a n = 8 1 n ( + 9 2) 4 ) n 3

13 Name-Surname: CLASSWORK 1 Find the number of integer solutions of the equation x 1 + x 2 + x 3 + x 4 + x 5 = 18 where 1 x i 10. Give 1 to each. Distribute the remaining 13 items to 5 boxes randomly. ( ) = 13 ( ) 17 = Now we have to subtract the number of distributions where one box gets 11 or more. Choose one, give it 10, distribute the remaining 3 items to 5 boxes randomly. ( )( ) The result is: = = 2205

14 Name-Surname: CLASSWORK 1 Find the number of integer solutions of the equation x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 24 where 2 x i 9. Give 2 to each. Distribute the remaining 12 items to 6 boxes randomly. ( ) = 12 ( ) 17 = Now we have to subtract the number of distributions where one box gets 10 or more. Choose one, give it 8, distribute the remaining 4 items to 6 boxes randomly. ( )( ) The result is: = = 5432

15 Name-Surname: CLASSWORK 2 There are three cards. 2 of them has FF written behind them, one of them has AA. There is a sentence on the front of each card: CARD I CARD II CARD III AA is here There is FF behind Card III AA is NOT here You know that of these 3 statements, at least one is TRUE and at least one is FALSE. Which card would you choose? (Assuming you want AA) There are three cases. The AA is behind Card I, or Card II or Card III. Card I Card II Card III Card I has AA T T T Card II has AA F T T Card III has AA F F F The first and third lines give impossible results. (All true or all false) Therefore second line is correct: Choose Card II for AA.

16 Name-Surname: CLASSWORK 3 Show that 4 5 n n + 1 for all n Z +. n = The claim is correct for n = 1. Assume it is correct for n = k, in other words, assume 5 k k + 1 = 4p Now consider n = k + 1: 5 k k = ) (4 5 k k + ( ) 5 k k + 1 = 4q + 4p = 4r where p, q, r are integers. Therefore the claim is correct by mathematical induction.

17 Name-Surname: CLASSWORK 3 Show that 8 3 2n n 1 for all n Z +. n = The claim is correct for n = 1. Assume it is correct for n = k, in other words, assume 9 k k 1 = 8p Now consider n = k + 1: 9 k k+1 1 = ) (8 9 k k + ( ) 9 k k 1 = 8q + 8p = 8r where p, q, r are integers. Therefore the claim is correct by mathematical induction.

18 Name-Surname: CLASSWORK 4 You have three containers. The two smaller ones have volumes of 64 and 59 cm 3. The third one is much larger. Its volume is around 1000 cm 3 but you don t know the exact number. You have unlimited amount of water. Can you put exactly 1 cm 3 of water to the large container? How? Adding once with 64 and removing once with 59 leaves 5 in the big container. Repeat this 12 times and you have 60. Now remove 59. A more systematic way is this: The greatest common divisor of 64 and 59 is 1, so it is possible. Using Euclid s algorithm: = = = = 0 Now reverse this process to obtain: 1 = = = So we have to add water 12 times using 64, and remove 13 times using 59.

19 Name-Surname: CLASSWORK 5 In a computer system, users have 4-letter passwords. Each character is a letter from the English alphabet, or a number. The system is not case sensitive. We are certain that if n the number of users then at least 5 users will have exactly the same password. What is n? Number of possible characters is: = 36 Number of possible passwords is: 36 4 = So the answer is: n = =

20 Name-Surname: CLASSWORK 6 In how many ways can we arrange the letters {A, a, B, b, C, c, D, d, E, e} as a 10-letter word such that lowercase and uppercase versions of the same letter are not next to each other? In total there are 10! permutations. Consider Aa as a unit (or Bb, or Cc etc.) Now there are 9 objects, so 9! permutations that we do NOT want. But considering Aa and aa as two separate items, there are 9! 2. Using inclusion exclusion, subtracting and adding, we obtain: 10! ( ) 5 9! ( ) 5 8! ( ) 5 7! ( ) 5 6! ( ) 5 5! =

21 Name-Surname: CLASSWORK 7 There are 9 students in a class. We want to form 5 project teams. Each team must have at least one member. In how many different ways can we do this? Hint: You may use Stirling table. Using Stirling table s values for S(8, 4) = 1701 and S(8, 5) = 1050 we obtain S(9, 5) = = 6951 OR Use inclusion-exclusion. Assign students to teams randomly. Then subtract the cases where one team is empty, then add the cases where two teams are empty etc. 5 9 ( ) ( ) ! ( ) ( ) = 6951 (We have to divide by 5! because teams are NOT distinct)

22 Name-Surname: CLASSWORK 7 Find the number of onto functions from the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} to the set B = {a, b, c, d, e}. Hint: You may use Stirling table. Using Stirling table s values for S(8, 4) = 1701 and S(8, 5) = 1050 we obtain S(9, 5) = = 6951 Now we have to multiply by 5! because each element on the set B is distinct: ! = OR Use inclusion-exclusion. Assign numbers to letters randomly. Then subtract the cases where nothing is mapped into letter a, add the cases where nothing is mapped into letter a or b, etc. 5 9 ( ) ( ) ( ) ( ) =

23 Name-Surname: CLASSWORK 8 5 students, Beyza, Cem, Mertcan, Sarper and Talha will share 29 pieces of identical chocolates according to the following constraints: Beyza gets at least 5. Cem gets at most 9. Mertcan gets at most 3. Sarper gets a multiple of 4. (including 0) In how many different ways can they do this? The series representing these constraints are: ( x 5 +x 6 +x 7 + )(1+x+x 2 + +x 9 )(1+x+x 2 +x 3 )( 1+x 4 +x 8 +x 12 + )( ) 1+x+x 2 +x 3 + = x5 1 x 1 x10 1 x 1 x4 1 x 1 1 x x = x5 x 15 (1 x) 4 ) ( ) n + 3 = (x 5 x 15 x n n n=0 The coefficient of x 29 is: ( ) ( ) 17 =

24 Name-Surname: CLASSWORK 8 5 students, Berkay, Gizem, Merve, Sena, and Yunus will share 37 pieces of identical chocolates according to the following constraints: Berkay gets an odd number. Gizem gets at least 7. Merve gets at most 5. Sena gets 0 or 1. In how many different ways can they do this? The series representing these constraints are: ( )( )( )( ) x+x 3 +x 5 + x 7 +x 8 +x 9 + )(1+x+ +x 5 1+x 1+x+x 2 +x 3 + = x 1 x 2 x 7 1 x 1 ( ) x6 1 x 1 + x 1 1 x = x8 x 14 (1 x) 4 ) ( ) n + 3 = (x 8 x 14 x n n n=0 The coefficient of x 37 is: ( ) ( ) 26 =

25 Name-Surname: CLASSWORK 9 Find the Θ( ) class (i.e. asymptotic complexity) of the following algorithm: INPUT: integer n S = 0 For i = 1 to n For j = i to i + 17 S = S + 1 EndFor EndFor Return S Number of operations is: 18 n, so asymptotic complexity is Θ(18n). But, a better way to express this fact is: Θ(n)

26 Name-Surname: CLASSWORK 9 Find the Θ( ) class (i.e. asymptotic complexity) of the following algorithm: INPUT: integer n S = 0 For i = 2 to 2 * n For j = i - 1 to i + 1 S = S + 1 EndFor EndFor Return S Number of operations is: (2n 1) 3, so asymptotic complexity is Θ(6n 3). But, a better way to express this fact is: Θ(n) Author: Dr. Emre Sermutlu June 23, 2016

First Midterm Examination

2015-2016 Fall Semester First Midterm Examination 1) 6 students will sit at a round table. Anıl, Sümeyye and Tahsin are in section 1 and Bora, İpek and Efnan are in section 2. They will sit such that nobody