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- Kansas
- University of Kansas
- Electrical Engineering
- Electrical Engineering 210
- Kinnersley
- Lecture 10: Fundamentals of Counting

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1 Lecture 10: Fundamentals of Counting Read: Chpt.5.1-5.5, 7.5, Rosen Consider that we are performing two experiments A and B with A having m possible outcomes a 1 , a 2 , ?, a m and B having n possible outcomes b 1 , b 2 , ?, b n . Q: What are the possible outcomes if we are to perform either A or B, but not both, and what are the possible outcomes if we are to perform both A and B? Possible Outcomes: ? Either A or B: {a 1 , a 2 , ?, a m , b 1 , b 2 , ?, b n }. # possible outcomes = |A| + |B| = m + n. ? Both A and B: {(a 1 ,b 1 ), (a 1 ,b 2 ), ?, (a 1 ,b n ), (a 2 ,b 1 ), (a 2 ,b 2 ), ?, (a 2 ,b n ), ? (a m ,b 1 ), (a m ,b 2 ), ?, (a m ,b n )}. # possible outcomes = |A| * |B| = m ? n. Q: What if we are performing k tasks (experiments) A 1 , A 2 , ?, A k such that there are n i ways (outcomes) to perform task A i , 1 ? i ? k? 2 1. Two Basic Counting Principles: The Rule of Sum: There are n 1 + n 2 + ? + n k ways to perform exactly one of these k tasks. The Rule of Product: There are n 1 ? n 2 ? ? ? n k ways to perform all of these k tasks. Applications: 1. A menu in a local deli offers 6 soda, 10 sandwiches, 12 desserts. #ways to order a soda: 6. #ways to order exactly one item: 6 + 10 + 12 = 28. #ways to order a full meal (soda + sandwiches + dessert): 6 ? 10 ? 12 = 720. #ways to order a soda and a sandwich: 6 ? 10. #ways to order any two items: 28 ? 28 = 784. #ways to order any two distinct items: 28 ? 27 = 756. #ways to order two items of different kind: (6 ? 10) + (10 ? 12) + (6 ? 12) = 252. 3 2. An electronic lock uses a 7-digit code to operate it. #possible codes: 10 7 . #possible codes starting with 0: 1?10 6 . #possible codes not starting with 0: 9?10 6 . (#possible codes ? #possible codes starting with 0 = 10 7 ? 10 6 = 9?10 6 .) #possible codes starting with 0, or 1, or 2: 3?10 6 . #possible codes starting and ending with 0: 1?10 5 ?1. #possible codes starting or ending with 0 but not both: (1?10 5 ?9) + (9?10 5 ?1) #possible codes starting with 1, 3, 5 or ending with 2, 8: 3?10 6 + 10 6 ?2 ? 3?10 5 ?2 #possible codes starting with 1, 3, 5 or ending with 2, 8 but not both: (3?10 5 ?8) + (7?10 5 ?2) 4 3. In Fortran, an identifier consists of a character string with 1 to 7 characters such that the first character must be a letter chosen from {a, b, ?, z}, followed by up to 6 more letter(s) and/or digits. Q: How many different identifiers are there? Let P be the # of all possible identifiers, P 1 be the # of all identifiers with exactly 1 char, P 2 be the # of all identifiers with exactly 2 chars, ? P 7 be the # of all identifiers with exactly 7 chars. P 1 = 26, // 26 ways to chose the 1 st char P 2 = 26?36, // 36 ways to chose the 2 nd char P 3 = 26?36 2 , // 36 2 ways to chose the 2 nd & 3 rd chars ? P 7 = 26?36 6 . // 36 6 ways to chose the 2 nd to 7 th chars ? P = P 1 + P 2 + ? + P 7 = 26?(36 0 + 36 1 + ? + 36 6 ) = 26?( 7 36 1 36 1 ? ? ) ? 5.82×10 10 . Practice HW: Chpt.5.1, 3, 5, 21, 23, 25, 27, 29, 31, 35. 5 2. The Pigeonhole Principle: Theorem: If k+1 objects (pigeons) are to be placed in k boxes (pigeonholes), then at least one of the k boxes must contain two or more of the given objects. Proof. (Proof by Contradiction) Assume that k+1 objects are placed in the k boxes but none of the k boxes contains more than one object to obtain a contradiction. Since each box contains at most one object, the maximum number of objects in these k boxes must be ? k, which contradicts to the assumption that there are k+1 objects in the k boxes. Hence, the Pigeonhole Principle holds. Remark: This is also known as Shoebox Principle, or Dirichlet Principle. Applications: 1. Among 13 people in a party, 2 of them must have their birthday in a same month. 2. Among 102 students taking an exam (max = 100 points), 2 of them must have same score. 3. Given 210 pairs of married couples (420 people). A minimum of 211 people must be invited so as to guarantee that at least a married couple will be included. 6 More Interesting Applications: 4. Given any m arbitrary positive integers a 1 , a 2 , ?, a m , m ? 1. Prove that there exist integers i, j, 0 ? i < j ? m, such that the sum a i+1 + a i+2 + ? + a j is divisible by m. Proof. Define the following partial sums: S 1 = a 1 , S 2 = a 1 + a 2 , ? S m = a 1 + a 2 + ? + a m . Consider the following two cases. Case 1: If ? k, 1 ? k ? m, such that m ?S k , we can choose i = 0 and j = k. Case 2: Assume that ? k, 1 ? k ? m, m ? S k . By Division Theorem, we must have S 1 = mq 1 + r 1 , 1 ? r 1 ? m ?1, S 2 = mq 2 + r 2 , 1 ? r 2 ? m ?1, ? S m = mq m + r m , 1 ? r m ? m ?1. Observe that there are m remainders r 1 , r 2 , ?, r m but only m ?1 values for them to choose from. Hence, by Pigeonhole Principle, two of the m remainders must be the same! WOLOG (WithOut Loss Of Generality), let r s = r t = r, and 1 ? s < t ? m. 7 Hence, we have S s = mq s + r, and S t = mq t + r. ? S t ? S s = m(q t ? q s ) and m ?S t ? S s . By construction, S t ? S s = (a 1 + a 2 + ?+ a s + ? + a t ) ? (a 1 + a 2 + ? + a s ) = (a s+1 + a i+2 + ? + a t ). Hence, m ?(a s+1 + a i+2 + ? + a t ) and we can then choose i = s and j = t. Since both cases lead to the conclusion that ? i, j, 0 ? i < j ? m, such that m ?(a i+1 + a i+2 + ? + a j ), the assertion must be true. 8 5. A chess grandmaster has 11 weeks to prepare for a tournament and he/she has decided to play at least 1 game each day but no more than 12 games during any 7 consecutive days. Prove that there must be a period of consecutive days during which the grandmaster will play exactly 21 games. Proof. There are 77 days in 11 weeks. Let a 1 be the total #games played on the first day, a 2 be the total #games played on the first two days, ? a 77 be the total #games played on the first 77 days. ? 1 ? a 1 < a 2 < ? < a 77 ? 11?12 = 132, and 22 ? a 1 +21 < a 2 +21 < ? < a 77 +21 ? 153. (Why?) Every one of the numbers {a 1 , a 2 , ?, a 77 , a 1 +21, a 2 +21, ?, a 77 +21 } must satisfy the following inequality: 1 ? a 1 , a 2 , ?, a 77 , a 1 +21, a 2 +21, ?, a 77 +21 ? 153. Since these 154 numbers are between 1 and 153, by Pigeonhole Principle, at least two of them must be the same. Hence, ? i, j ? N, 1 ? i < j ? 77, such that a j = a i + 21. (Why?) Hence, a j ? a i = 21 and, from day (i+1) to day j, the master must have played exactly 21 games. 9 6. Consider the set S = {1, 2, ?, 2n}. Prove that among any arbitrarily chosen n+1 integers from S, there must exists a pair of integers x and y such that x ?y. Proof. Let x 1 , x 2 , ?, x n+1 be any n+1 integers arbitrarily chosen from S. By factoring out as many 2?s from each integer, each integer x i , 1 ? i ? n+1, can be written as: x i = 2 ki q i , where k i ? 0 is an integer and q i is an odd integer with 1 ? q i ? 2n?1. Examples: 28 = 2 2 7, 39 = 2 0 39. Consider these n+1 odd integers q 1 , q 2 , ?, q n+1 . Since q i is an odd integer with 1 ? q i ? 2n?1 and there are at most n distinct values for the q i ?s, ? i, j, 1 ? i < j ? n+1, such that q i = q j = q. Consider the integers x i = 2 ki q and x j = 2 kj q. WOLOG, let x i < x j . Hence, k i < k j . (Why?) ? x j = 2 kj q = (2 ki 2 (kj?ki) )q = (2 ki q)2 (kj?ki) = x i 2 (kj?ki) Since 2 (kj?ki) ? N, we have x i ?x j . By choosing x = x i and y = x j , we have the assertion. 10 Extension: Generalized Pigeonhole Principle: If m objects are to be placed in n boxes, m ? n, then there exists at least one box containing at least ? n m ? objects. Special cases: 1. If m = n + 1, we have the Pigeonhole Principle. 2. If m = kn + 1, then there exists at least one box containing at least k + 1 objects. Applications: 1. If there are 37 people in a party, at least how many of them were born in the same month? Since m = 37, n = 12, # people born in the same month ? ? 37 12 ? = 4. 2. At least how many guests you must invite in order to guarantee that 6 of them were born in the same month? Since ? n m ? = 6 with n = 12, m = 61. Hence, at least 61 guests must be invited. Another Approach: Since n = 12 and k + 1 = 6 (or k = 5), m = kn + 1 = 5?12 + 1 = 61. 11 3. Assume that there are infinitely that many red, white, and blue socks in a laundry basket. Q: How many socks one must select to guarantee 3 pairs of socks of the same color? Since n = 3, k + 1 = 6 (or k = 5), m = kn + 1 = 5?3 + 1 = 16. Hence, at least 16 socks must be selected. Q: How many socks one must select to guarantee 3 pairs of red socks? Infinity! 4. Prove that in a party of six, either 3 of them are mutual friends or 3 of them are complete strangers to each other. Proof. Let A be any person in the group. Among the 5 remaining people, by Generalized Pigeonhole Principle, either 3 or more of them are friends of A, or 3 or more of them are strangers to A. Case 1: Let B, C, D be friends of A. If two of them are friends, together with A, we have 3 mutual friends. Else, they are 3 complete strangers to each other as required. Case 2: Let B, C, D be strangers to A. If 2 of them are strangers to each other, together with A, we have 3 complete strangers. Else, they are 3 mutual friends as required. Practice HW: Chpt.5.2, 3, 7, 9, 13, 17, 19, 21, 25. 12 3. Permutation and Combination: Let?s consider the ordered arrangements of three distinct objects a, b, and c. They are: abc acb bac bca cab cba Each one of these arrangements is a permutation of {a,b,c}. Dfn: Given a set of n distinct objects S. An r-permutation of S, r ? n, is an ordered arrangement (selection/placement) of any r elements of S. Let P(n,r), n r P , be the # of r-permutations of n objects. Example: Let S = {a,b,c}. P(3,1) = 3, P(3,2) = 6, P(3,3) = 6. 13 Q: How do we compute P(n,r)? Observe that we have n choices for the 1 st object, n?1 choices for the 2 nd object, n?2 choices for the 3 rd object, ? n?r+1 choices for the r th object. Hence, P(n,r) = n(n?1)(n?2)?(n?r+1) = (n) r , (falling factorial function) = ! ()! n nr? . Observe that P(n,n) = n!, P(n,0) = 1, P(0,0) = 1. Applications: 1. In how many different ways can a 5-character string be formed from {a, b, c, d, e, f, g}? P(7,5) = 7! 2! = 7?6?5?4?3 = 2,520 14 2. There are 30 members in a social club. In how many different ways can a committee with 1 chairperson, 1 vice-chairperson, 1 secretary, and 1 treasurer be formed? P(30,4) = 30! 26! = 657,720 3. Given S = {2, 3, 5, 7, 9}. (a) How many distinct 3-digit integers can be formed? P(5,3) = 5! 2! = 60 (b) How many of these integers are less than 500? 2 ways to choose the 1 st digit, 4 ways to choose the 2 nd digit, 3 ways to choose the 3 rd digit, ? # distinct 3-digit integers < 500 = 2?4?3 =24 (c) How many of these distinct 3-digit integers are odd? 3?4?4 =48 (d) How many 3-digit integers < 500 are odd? 24 ? 3 = 21 (Why?) 15 Let?s now consider the unordered arrangements of distinct objects. Dfn: Given a set S of n distinct objects. An r-combination of S is an unordered arrangement (selection/placement) of r. Let C(n,r) = n r C = () n r be the # r-combinations of n objects. Example: Let S = {a,b,c}. 1-combination of S: a, b, c 2-combination of S: ab, ac, bc 3-combination of S: abc C(3,1) = 3, C(3,2) = 3, C(3,3) = 1. 16 Q: How do we compute C(n,r)? Theorem: P(n,r) = C(n,r)?P(r,r). Proof. To orderly arrange r objects in S, we may first select, without regarding to order, any r objects and then orderly arrange these r objects in all possible ways. Hence, P(n,r) = C(n,r)?P(r,r). Corollary: C(n,r) = (,) (,) Pnr Prr = ! ()!! n nrr? = C(n,n-r). Applications: 1. In how many different ways can a committee of 5 be formed among 30 members? C(30,5) = 30! 25!5! = 142,506 2. A menu has 6 sodas, 10 sandwiches and 5 desserts. In how many different ways can we order 3 sodas, 5 sandwiches and 2 desserts? C(6,3)?C(10,5) ?C(5,2) = 50,400 17 3. A student must answer 10 out of 13 questions in an exam. (a) How many different ways can he/she take this exam? C(13,10) = 286 (b) What if he must answer the first 2 questions? C(11,8) = 165 (c) What if he must answer either the first or second, but not both, questions? C(2,1)?C(11,9) = 110 (d) What if he must answer exactly 3 out of the first 5 questions? C(5,3) ?C(8,7) = 80 (e) What if he must answer at least 3 out of the first 5 questions? C(5,3) ?C(8,7) + C(5,4) ?C(8,6) + C(5,5) ?C(8,5) = 276 4. Pascal Identity: C(n,r) = C(n-1,r) + C(n-1,r-1). Proof. Let x be any object in S. Any selection of r objects from S will either include or exclude x. If x is included, #ways to select the remaining r-1 objects is C(n-1,r-1). If x is excluded, #ways to select the r objects is C(n-1,r). Hence, by Sum Rule, we have C(n,r) = C(n-1,r) + C(n-1,r-1). Practice HW: Chpt.5.3, 13, 15, 17, 21, 25, 27, 33, 39. 18 4. The Principle of Inclusion and Exclusion: For finite sets A, B, |A ? B| = |A| + |B| ? |A ? B|. For finite sets A, B, C. |A ? B ? C| = |A| + |B| + |C| ? |A ? B| ? |B ? C| ? |A ? C| + | A ? B ? C|. ? In general, for finite sets A 1 , A 2 , ?, A n , |A 1 ? A 2 ? ? ? A n | = ? 1?i?n |A i | ? ? 1?i n, there is no injection from A to B. (f) If m ? n, there are (n) m injections from A to B. Q: How many surjections are there from A to B? (g) If m < n, there is no surjection from A to B. 23 (h) If m ? n, the number of surjections from A to B is: 1 ( ,1)( 1) ( ,2)( 2) ... ( 1) ( , 1)1 . mm mnm nCnn Cnn Cnn ? ??+??+? ? Proof. Let B = {b 1 , b 2 , ?, b n }. If a function f from A to B is not a surjection, one or more b i ?s must not have a pre-image in A. Define: P 1 be the property that b 1 has no pre-image in A, P 2 be the property that b 2 has no pre-image in A, ? P n be the property that b n has no pre-image in A. Hence, # of surjections = N(P 1 ?P 2 ??P n ?) = n m ? ? 1?i?n N(P i ) + ? 1?i

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Textbook: Discrete Mathematics: Mathematical Reasoning and Proof with Puzzles, Patterns, and Games

Created: 2009-01-24

Updated: 2009-01-24

File Size: 44 page(s)

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Textbook: Discrete Mathematics: Mathematical Reasoning and Proof with Puzzles, Patterns, and Games

Created: 2009-01-24

Updated: 2009-01-24

File Size: 44 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 11

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