184 DISCRETE MATHEMATICAL STRUCTURES 7 Elementary Combinatorics 7.1 INTRODUCTION Combinatorics deals with counting and enumeration of specified objects, patterns or designs. Techniques of counting are important in mathematics and computer science. In this chapter, we shall study basic of counting. We also develop basic ideas of permutations, combinations, Binomial theorem, power sets and pigeon hole principle. We begin our study with two basic counting principles. 7.2 BASICS OF COUNTING 7.2.1 Sum Rule (Principle of Disjunctive Counting) Let S be a set and S denote the number of elements in S. If S is a union of disjoint non-empty subsets A 1, A 2,, A n then S = A 1 + A 2 + + A n In the above statement the subsets A i of S are all disjoint i.e., they have no element in common. If A i and A j an two subsets of S, then A A = ( for i j) and we have i j S = A1 A2 A3... An that is each element of S is exactly in one of the subsets A i. In other words, the subsets A 1, A 2, A n is a partition of S. We now state the sum rule for counting events. Let S be a sample space. Two events E 1 and E 2 of X are said to be mutually exclusive if the events have no elements in common. If E 1, E 2, E 3, E n, are mutually exclusive events of S, then we can state sum rule for counting events as follows: If E 1, E 2, E n are mutually exclusive events of a sample space S and E 1 can happen in m 1 ways. E 2 can happen in m 2 ways,, E n can happen in m n ways then E 1 or E 2 or or E n can happen m 1 + m 2 + + m n ways. Example 1: How many ways can we get a sum of 7 or 1 when two distinguishable dice are rolled? Solution: The two dice are distinguishable, therefore the ordered pairs (a, b) and (b, a) are distinct when a b, i.e., ( a, b) ( b, a) for a b. The ordered pairs in which the sum is 7 are:
ELEMENTARY COMBINATORICS 185 (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). These ordered pairs are distinct. There are 6 ways to obtain the sum 7. Similarly the ordered pairs: (5, 6), (6, 5) are all distinct. The number of ways in which we get a sum 11 with the two dice is 2. Therefore, we can get a sum 7 to 11 with two distinguishable dice in 6 + 2 = 8 ways. Example 2: How many ways can we draw a club or a diamond from a pack of cards? Solution: There are 13 clubs and 13 diamonds in a pack of cards. The number of ways a club or a diamond may be drawn 13 + 13 = 26. Example 3: In how ways can be drawn an ace or a king from an ordinary deck of playing cards? Solution: Number of Aces in a pack = 4 Number of kings in a pack = 4 Number ways an Ace or a king can be drawn from the pack = 4 + 4 = 8 7.2.2 Product Rule (The Principle of Sequential Counting) If A 1, A 2,, A n are non-empty sets, then the number of elements in the Cartesian product A 1 A 2 A n is the product A n i= 1 i n A A... A = A 1 2 Product rule in terms of events: If E 1, E 2,, E n are events, and E can happen in m 1 ways E 2 can happen m 2 ways, E n can happen in m n ways, then the sequence of events E, first followed by E 2 followed by E 3,, followed by E n can happen in m 1 m 2 m n ways. Example 1: How many possible out comes are there when we roll a pair of dice, one red and one green? Solution: The red die can land in any one of six ways and for each of there six ways, the green die can also land in six ways. The number of possible out comes when two dice are rolled = 6 6 = 36. Example 2: In how many different ways one can answer all the questions of a true-false test consisting of 4 questions? Solution: There are two ways of answering each of the 4 questions. Therefore by product rule the number of ways in which all the 4 questions can be answered. = 2 2 2 2 = 16 Example 3: Find the number n of license plates that can be made where each plate contains two distinct letters followed by three different digits. Solution: First letter can be printed in 26 different ways. Since the second letter must be different from the first, we have 25 contains for the second letter. Similarly the first digit can be printed in 10 ways, the second digit in the license plate can be printed in 9 ways and the third in 8 ways. n i= 1 i
186 DISCRETE MATHEMATICAL STRUCTURES Therefore, the number of license plates that can be printed, so that each plate contains two distinct letters follower by three different digits 26. 25. 10. 9. 8 = 4,68,000. E XERCISE 7.1 1. In how many ways can be draw a king or a queen from ordinary deck of playing cards? 2. How many ways can we draw a club or a spade from a pack of cards? 3. How many ways can we get a sum 6 or 9 when two distinguishable dice are rolled? 4. How many possible out comes are there when we roll a pair of dice, one yellow and one red? 5. In how many different ways are can answer all the questions of a true-false test consisting of 5 questions? 6. How many ways can we get a sum of 8 when two in distinguishable dice are rolled? 7. How many ways can we get a sum of 4 or 8 when two distinguishable dice are rolled? How many ways can we get an even sum? 8. In how many ways can an organisations containing 26 members elect a president, a treasurer and a secretary (assuming no person is elected to more than one position)? 9. In a railway compartment 6 seats are vacant on a bench. In how many ways can 3 passengers can sit on them? 10. There are 10 buses plying between on a bench. In how many ways can 3 passengers can sit on them? 11. Suppose a license plate contains two letters followed by three digits with the first digit not zero. How many different plates can be printed? 12. Suppose a license plate contains 3 English letters followed by 4 digits: (a) How many different license plates can be manufactured if repetition of letter and digits are allowed? (b) How many plates are possible if only the letters are repeated? (c) How many are possible if only the digits can be repeated? 13. How many different license plates are there that involve 1, 2 or 3 letters followed by 4 digits? 14. There are 10 true-false questions on an examinations. How many sequences are possible? 15. Suppose that a state s license plates consists of three letters followed by three digits. How many different plates can be manufactures (if repetitions are allowed)? 16. If there are 12 boys and 16 girls in a class find the number of ways of selecting one student as class representative. Answers: 1. 8 2. 26 3. 9 4. 36 5. 32 6. 12 7. 8, 18 8. 15,600 9. 120 10. 90 11. 608400 12. 26 3 10 4, 26 3 10 9 8 7, 26 25 24 10 4 13. (26 + 26 2 + 26 3 ) 10 4 14. 1024 15. 26 3 10 3 16. 28.
ELEMENTARY COMBINATORICS 187 7.3 PERMUTATIONS AND COMBINATIONS 7.3.1 Permutations Definition 7.1: A permutation of n objects taken r at a time is an arrangement of r of the objects (r n). A permutation of n objects taken r at a time is also called r-permutation or an r-arrangement. Various symbols are used to indicate the number of permutations of n things taken r at a time. The one we shall use in this text is P (n, r), (r n). The symbols n Pr or P ( n r ), [n] r or n (r) are also used to denote the number of permutations of n objects taken r at a time. Example 1: Consider the three letters a, b, c. The arrangements of the letter a, b, c taken two at a time are. ab, ba, ac, ca, bc, cb The number of 2-arrangements are 6 i.e., the number of permutation of 3 letters taken 2 at a time = 3 = P( 3, 2) = 6. P2 Notation In this chapter, we use {P 1 a, P 2 b, P 3 c, P 4 d} to indicate either: (i) That we have P 1 + P 2 + P 3 + P 4 objects which include P 1 a s, P 2 b s, P 3 c s, P 4 d s. (ii) That we have 4 objects a, b, c and d with the condition that: a can be chosen at most P 1 times. b can be chosen at most P 2 times. c can be chosen at most P 3 times. d can be chosen at most P 4 times. The numbers P 1, P 2, P 3 and P 4 are called repetition numbers. Example 2: The three permutations of {3 a, 1 b, 1 c} are aaa, aab, baa, aac, aca, abc, acb, bac, bca, cab, cba, aba. 7.3.2 Factorial Function Definition 7.2: If n is a natural number, then the product of all the natural numbers from 1 to n is called n-factorial. It is denoted by the symbol The symbols and n (n) are also used. From the definition. = n (n 1) (n 2) 3.2.1 The factorial function can also be defined recursively as follows: 0! = 1 (n + 1)! = (n + 1), n 0 From the above recursive definition, we get 1! = 0! (1) = 1 2! = 1!(2) = 1. 2 3! = 2!(3) = 1.2.3
188 DISCRETE MATHEMATICAL STRUCTURES 7.3.3 Falling and Rising Factorials Definition 7.3: The number nn ( 1) ( n 2)...( n r + 1) is called falling factorial. It is denoted by [n] r [n] r The number nn ( + 1) ( n+ 2)... ( n+ r 1) is called rising factorial. It is denoted by the symbol 7.3.4 Stirling Numbers Definition 7.4: as The falling factorial [x] r is a polynomial of rth degree is x. Let us unite the polynomial 0 1 2 2 L r r r r r r r [ x] = S + S x + S x + + S x The numbers 0 1 2 r r, r, r,..., r S S S S are called the stirling numbers of the first kind. We have The recurrence relation is 0 r k r r r S = 0, S = 1, S = 0, if k > r k k 1 k 0 r r r r r r S + 1 = S r S, S = 0, S = 1 Let n r the number of distributions of n distinct objects into r non-distinct, boxes, no box being r empty is denoted by S n. It is called a stirling number of the second kind. We have the recurrence formula. The number 1 2 + +... + n n n n r r 1 r n n n S + 1 = S + r S,if1< r < n 1 n S1 = 1, Sn = 1, Sn = 0 if r > n 0 r 0 0 0 n S = 1, S = S = 0 if r, n > 0 S S S is called nth Bell number. It is denote by B n Theorem 7.1: (Number of r-permutation without repetition). The number of r-permutations of n objects r! is p (n, r) = n (n 1) (n 2) (n r+1) = ( n r) Proof: The first element in an r-permutations of n objects can be chosen in n different ways. Once the first element has been selected, the second element can be selected in n 1 ways. We continue selecting the elements, having selected (r 1)th element. We select rth element i.e., the last element. The rth element in the permutation can be selected in n r +1 ways. By product rule, the number of r-permutations of a set of n distinct objects is P (n, r) = n (n 1) (n 2) (n r +1) We may also write P (n, r) in factorial
ELEMENTARY COMBINATORICS 189 P (n, r) = n (n 1) (n 2) (n r +1) = =! ( n r) ( ) ( )( ) ( n r)... 2 1 n n 1... n r + 1 n r... 2 1 7.4 SOLVED EXAMPLES Example 1: Find 8! Solution: For n 0, we have Hence, (n + 1) = (n + 1) 8! = (7 + 1)! = 7! 8 = 6! 7 8 = 5! 6 7 8 = 4!5 6 7 8 = 3!4 5 6 7 8 = 2!3 4 5 6 7 8 = 1!2 3 4 5 6 7 8 = 1 2 3 4 5 6 7 8 = 40,320 Example 2: Simplify ( n + ) Solution: or 1! ( n ) ( n ) n( n ) n n( n ) + 1! + 1 1...3 2 1 = = n + 1! 1... 3 2 1 ( + ) ( + ) n 1! n 1 = = + Example 3: Prove that 2 P (n, n 2) = P (n, n) Solution: ( n 1) = 2 P (n, n 2) = 2 ( n ( n )) 2 = 2 = n n + 2! 2! = = P (n, n). 2!
190 DISCRETE MATHEMATICAL STRUCTURES Example 4: How many words of three distinct letters can be formed form the letters of the word LAND? Solution: The number of three distinct letter words that can be formed from the 4 letters of the word LAND is P (4,3) = ( ) 4! 4! = = 4! = 24 4 3! 11 7.5 PERMUTATIONS WITH LIKE ELEMENTS Theorem 7.2: are alike is The number of permutations of n objects of which are q 1 are alike, q 2 are alike,, q r P (n, q, q 2,, q r) = q! q!... q! 1 2 r Where n = q 1 + q 2 + + q r Proof: Let the number of permutation be x. If the q 1 like objects are unlike, then for each of these x arrangements, the q 1 like objects could be rearranged among themselves in q 1! Ways without altering the positions of the other objects. The number of permutations would be x q 1! Similarly; if q 2 like objects were unlike; each of these x q 1! Permutations would give rise to q 2! Permutations Therefore, the number of permutations would be x q 1! q 2! Similarly, if all the objects were unlike number of permutations would be x q! q 2! q r! But if all the objects were unlike, the number of permutations with the n objects would be Hence xq! q 2! q r! = We have x = q1! q 2!... qr! i.e., P (n q 1, q 2,, q r ) = q1! q 2!... qr! Example: There are 4 black, 3 green and 5 red balls. In how many ways can they be arranged in a row? Solution: Total number of balls = 4 black + 3 green + 5 red = 12 The black balls are alike, The green balls are, and the red balls are alike, The number of ways in which the balls can be arranged in a row = 7.5.1 Ordered Samples 12! 27, 720 4!3!5! = In combinational analysis, many problems are concerned with choosing a ball from an urn containing n balls or a card from a deck. When we choose one ball after the other from the urn (or a card from a deck) say r times, we call each choice an ordered sample of size r, we have two cases:
ELEMENTARY COMBINATORICS 191 (i) Sampling with replacement The drawn ball may be replaced in the urn before the next ball is drawn. There are n different ways to choose each ball, therefore by fundamental principle of counting (product rule); there are n r different ordered sample with replacement of size r. (ii) Sampling without replacement When the ball drawn is not replaced in the urn after it is drawn; repetitions do not occur, in the ordered sample. In this case, an ordered sample of size r without replacement, is an r-permutation of the objects in the urn. Hence there are ( n r)! different ordered samples of size r without replacement. Example 1: Suppose an urn contain 5 balls. Find the number of ordered sample of size 2. (i) With replacement (ii) Without replacement Solution: (i) There are 5 balls, and each ball in the ordered sample can be chosen in 5 ways. Hence, there are 5.5 = 25; samples with replacement. (ii) The first ball in the ordered sample can be chosen in 5 ways and the next ball in the ordered sample can be chosen in 4 ways (when the first drawn ball is not replaced). There are 5 4 = 20, samples without replacement. Example 2: In how many ways can one choose two cards in succession from a deck of 52 cards, such that the first chosen card is not replaced. Solution: There are 52 cards in the deck of cards since the chosen card is not replaced the first can be chosen in 52 different ways and the second can be in 51 different ways. The number of ways in which the 2 cards are chosen = 52 51 = 2,652 Example 3: A box contains 10 light bulbs. Find the number n of ordered samples of: (a) Size 3 with replacement, and (b) Size 3 without replacement. Solution: (a) n = 10 r = 10 3 = 10 10 10 = 1,000 and (b) P (10, 3) = 10 9 8 = 720 7.6 CIRCULAR PERMUTATIONS Definition 7.5: A circular permutation of n objects is an arrangement of the objects around a circle. In circular arrangements, we have to consider the relative position of the different things. The circular permutations are different only when the relative order of the objects is changed otherwise they are same.
192 DISCRETE MATHEMATICAL STRUCTURES 7.6.1 Number of Circular Permutations Fig. 7.1 Definition 7.6: Let n distinct be given. If the n objects are to be arranged round a circle we take an objects and fix it in one position. Now the remaining (n 1) objects can be arranged to fill the (n 1) positions the circle in (n 1)! ways. Hence the number of circular permutations of n different objects = (n 1)! 7.6.2 Number of Different Circular Permutations We consider the order; clockwise or anti-clockwise of objects around a circle as the same circular permutation. Every arrangement with n objects round a circle is counted twice in (n 1)! circular permutations. The total number of different permutations of n distinct objects is ( n ) 1! = 2 Example 1: In how many ways can a party of 9 persons arrange themselves around a circular table? Solution: One person can sit at any place in the circular table. The other 8 persons can arrange themselves in 8! ways i.e., the 9 persons can be arranged among themselves round the table in (9 1)! = 8! ways. Example 2: In how many ways 5 gents and 4 ladies dine at a round table, if no two ladies are to sit together? Fig. 7.2 Solution: Since no two ladies are to sit together, they should, seat themselves in between gents (i.e., a lady is to be seated in between two gents). The 5 gents can sit round the circular table in 5 positions (marked G in the Fig. 7.2). They can be arranged in (5 1)! = 4! ways. The ladies can sit in the 4 out of 5 seats (marked X in Fig. 7.2). This can be done in P (5, 4) ways. The required number of ways in which 5 gents and 4 ladies can sit round a table. 4! P 5, 4 = ( ) = ( 4 3 2 1) ( 5 4 3 2) = 2,880
ELEMENTARY COMBINATORICS 193 Example 3: Twelve persons are made to sit around a round table. Find the number of ways they can sit such that 2 specified are not together. Solution: 12 persons can sit round a table in (12 1)! = 11! ways. The total number of ways in which 2 specified persons are together is 2! 10!. The required number of seating arrangements in which 2 specified persons are not together. = 11! 2! 10! = 11 10! 2! 10! = 10! (11 2) = 9 10! E XERCISE 7.2 1. In how many ways can 5 Telugu, 3 English, 2 Hindi books arranged on a shelf, if the books of each language are to be together? 2. Suppose a license plate contains two letters followed by three digits with the first digit not zero. How many different license plates can be printed? 3. How many license plates can be formed involving 3 English letters and 4 non-zero digits, if all the letters must appear either in the beginning or in the end. 4. From the digits 1, 2, 3, 4, 5, 6 how many three digit odd numbers can be formed, if the repetition of digits is not allowed? 5. There are 10 true-false questions on an examinations. How many sequences are possible? 6. Suppose that a state s license plates consists of three letter followed by three digits. How many different plates can be manufactured (if repetitions are allowed)? 7. Suppose a license plate contain 1 or 2 letters followed 3 digits. How many different license plates can be printed? 8. Suppose a license plate contains 3 English letters followed by 4 digits. (a) How many different plates can be manufactured if repetition of letters and digits are allowed? (b) How many plates are possible if only the letters are repeated? (c) How many are possible if only the digits can be repeated? ( n + ) 9. Solve for n in ( n + ) 10. Find n if ( n ) 2! = 56 1! = 12 1! 11. A man wished to travel from one point in a city to a second point which is five blocks south and six blocks east of his starting point. In how many ways he make the journey if be always travels either south or east? 12. Find the number of ways of arranging the letter of the word TENNESSEE all at a time (a) if there no restriction (b) if the first two letters must be E. 13. Suppose a license plate contains three distinct letters followed by four digits with first digit not zero. How many different license plate can be printed?