1 Section 2.3 Purpose of Section To introduce some basic tools of counting, such as the multiplication principle, permutations and combinations. Introduction If someone asks you a question that starts off how many ways, you can be rest assured it s a problem in combinatorics. Someone once said that every mathematician is part combinatorist since combinatorial ideas and arguments are important in every area of mathematics, pure and applied. Today, the interest in combinatorial analysis is fueled by important problems in science, including chemistry, biology, physics and computer science. Problems in chemistry arise in studying the arrangement of atoms in molecules and crystals, in physics in the study of statistical and quantum mechanics, in biology in studying gene structure, and in computer science in a myriad of problems such as computer networks and microcircuit design. The general public too is familiar with combinatorics through games and puzzles, asking questions like how many games are played in the annual NCAA men s or women s basketball tournament? Sometimes it is easy to construct one way to do something, but hard to find the total number of ways to do the thing. For example, it is relatively easy to cover the 64 squares of an ordinary 8 8 checkerboard with dominoes (each domino covering 2 squares), but hard to determine the number of ways to do it. The total number of different coverings was discovered in 1961 by M.E. Fischer to be 2 4 2 901 12, 988,816 =. Although it is hard to formally define such a multifaceted subject, combinatorics is simply the discipline of counting. Counting is so commonplace, we generally don t give it a second thought, yet as the reader will see, the techniques of counting are as ingenious as any in mathematics and just as difficult. More than one good mathematician has been embarrassed by a seemingly simply combinatorics problem. Multiplication Principle One of the most basic principles of counting, if not the most basic, is the multiplication principle. Although the principle is simple, it has far reaching consequences. For example, how many dots are there in the array of dots in Figure 1?
2 Count the Dots Figure 1 We suspect you counted the 10 dots in the first row and then multiplied by 3. If this is true, then you used the multiplication principle. As a small step up in complexity chain, try counting the number of paths from A to C in Figure 2. No doubt this problem didn t stump you either, getting 4 5 = 20. Again you used the multiplication principle. How Many Paths from A to C? Figure 2 This leads us to the formal statement of the multiplication principle. Multiplication Rule for Counting: If a procedure can be broken into successive stages, and if there are s 1 outcomes for the first stage, s 2 outcomes for the second stage,, and s n for the n th stage, then the entire procedure has s 1 s 2..., s n outcomes.. Example 1 (Counting Subsets containing n elements has 2 n Subsets) Show that that a set A = { a a a } subsets.,,..., n Proof: A subset of A can be formed in n successive steps. On the first step one can pick or not pick a 1, on the second step one can pick or not pick a 2, and so on. One each step there are two options, to pick or not pick. Hence, the number of subsets that can be selected is 2 2 2 = 2 n 1 2
3 Example 2: (Counting Functions 1 ) How many functions are there from the set A a, b, c B = 0,1? Enumerate them. = { } to the set of binary numbers { } Solution For each of the 3 values in A the function can take on 2 values. 3 Hence, by the multiplication rule the number of functions is 2 2 2 = 2 = 8. We leave it to the reader to draw the eight functions (See Problem 21.) In general the number of functions from a set with cardinality n to a set with cardinality n m is m. Although the multiplication principle is very simple, it has far reaching results. One is the study of permutations. Permutations tions A permutation is simply an arrangement of objects. For example the permutations of the three letters abc are the six arrangements abc, acb, bac, bca, cab, cba The number of permutations increases dramatically as the size of the set increases. The number of permutations of the first 10 letters of the alphabet abcdefghij is 3,628,800. We certainly didn t arrive at that number by listing each arrangement. We used the multiplication principle. Suppose four individuals a, b, c, d are in a foot race and we wish to determine the possible ways the runners can finish first and second. Each of the four runners can finish first, and for each winner, there are 3 second-place finishers. Hence by the multiplication principle there are 4 3 = 12 possible ways the runners can finish first and second, which are. ab ba ca da ac bc cb db ad bd cd dc This leads to the definition of permutations of different sizes. Definition: A permutation of any r elements taken from a set of n elements is an arrangement of the r elements. We denote the number of such P n, r. permutations by 1 We will talk more about functions in Chapter 4.
4 Using the multiplication principle we can find the number of such permutations. Theorem 1 Number of Permutations The number of permutations of r elements taken from a set of size n is n! P ( n, r) = n n 1 n 2 n r 1! = + ( n r) Proof: Choosing r elements from a set of size n, we have the first element can be selected n ways. the second element can be selected n 1 ways (since now There are n 1 left). the third element can be selected n 2 ways (since now there are n 2 left). the r th element can be selected n r + 1 ways Hence, by the principle of sequential counting (or the multiplication rule), we have (, ) = ( 1) ( 2) ( + 1) P n r n n n n r. Example 3 Permutations of a Set with 3 Elements Find the permutations of 1, 2 and 3 Solution r = elements selected from {,, } a b c. The underlying set { a, b, c } has n = 3 elements. The permutations of 1, 2 and 3 elements from { a, b, c } are listed in Table 1.
5 r = 1 r = 2 r = 3 a ab abc b ac acb c ba bac bc bca ca cab cb cba P n, r Permutations Table 1 We don t use set notation for writing permutations 2 since order is important. The permutation ab is not the same as ba. n Factorial When counting sets, one often encounters the product of consecutive integers from 1 to n. This product is called n factorial and denoted by n! = n n 1 n 2 2 1 It is convenient to define 0! = 1. Also, note the number of permutations of n elements is P n, n = n n 1 n 2 2 1 = n! Margin Note: Permutations: Order matters. Margin Note: To evaluate (, ) P n r start at n and multiply r factors. P 4, 2 = 4 3 = 12 P 7,3 = 7 6 5 = 210 P 4,1 = 4 = 4 P 10,3 = 10 9 8 = 720 P 4, 4 = 4 3 2 1 = 24 2 Sometimes permutations are written with round parenthesis, such as ab is written as ( ab ).
6 Example 4 How many ways can one arrange the seven letters of the word SYSTEM? Solution The two S s are indistinguishable so we find the arrangements of the four letters Y,T,E,M taken from a set of size six and let the two S s occupy the remaining slots. Hence, we have 6! P ( 6, 4) = = 6 5 4 3 = 360 2! Another way to think of this problem is to momentarily distinguish the two S ' s as S 1 and S 2. In this way there are 6! permutations of the six letters. However, there are 2! = 2 permutations of the S 1 and S 2 so we must divide the 6! permutations by 2! getting the same result. A second application of the multiplication principle is computing combinations. Combinations Combinations are essentially permutations where order doesn t matter. For example, the combinations (or subsets) of size 2 that can be selected from the 3 letters in the word cat are { ca},{ ct},{ at } Note that there are fewer combinations than the 6 permutations of the letters cat. Definition: A combination is a subset of elements of a set. Example 5 Find the combinations (i.e. subsets) of size r = 1, 2 and 3 taken from the set { a, b, c }. Solution It is a simple matter to enumerate the combinations which are listed a, b, c with the exception of the empty Table 2. Note that the 8 subsets of { } set are listed. If we wanted a subset of size 0 we would include the empty set.
7 r = 1 r = 2 r = 3, a, b, c { a } { a b } { } { b } { a, c } { c } { b, c } Nonempty Subsets of { a, b, c } Table 2 Note: Note that combinations are simply sets, so { a, b } is the same as the combination { b, a }. Note too how this contrasts with permutations where the permutation ab is not the same as ba. Theorem 2: Number of Combinations The number of combinations (or subsets) of size r which can be selected from a set of size n, denoted by n C ( n, r) or r Proof: C ( n, r), is n n! = =. r r! ( n r)! Recall that the number of ways to permute r elements taken from a set n! P n, r =. But these r elements can be arranged in r! n r! of size n is different ways, so if we only want to count one of these permutations to obtain the number of combinations, we divide P( n, r ) by r! getting (, ) C n r P n, r n! = =. r! r! n r! ( ) n The numbers C ( n, r) or are called binomial coefficients because they are r the coefficients in the binomial expansion
8 2 2 2 2 2 2 2 2 a + b = a + ab + b = a + 2ab + b 0 1 2 3 3 3 3 3 3 2 2 3 3 2 2 3 a + b = a + a b + ab + b = a + 3a b + 3ab + b 0 1 2 3 4 4 4 4 4 a + b = a + a b + a b + ab + b = a + 4a b + 6a b + 4ab + b 0 1 2 3 4.................. 4 4 3 2 2 3 4 4 3 2 2 3 4. n It helps in thinking about combinations to say as "n choose r " since it r denotes the number of ways one can choose r items from a set of n items. 4 For example = 6 is read as 4 choose 2 is 6 which means there are 6 2 ways to choose 2 things from 4 things. Margin Note: Combinations: Order does not matter Example 6 How many ways can 10 players choose up sides to play fiveon-five in a game of basketball? Solution As in many combinatorial problems, there is more than one way to carry out the counting. Perhaps the simplest is to determine the number of ways one fixed player can select his or her four teammates from the 9 other players. In other words, the number of subsets of size four taken from a set of size 9, or nine choose four, which 9 9! 9 8 7 6 = = = 126 ways. 4 4!5! 4 3 2 1 Another approach is to find number of subsets of size 5 taken from a set of size 10 and divide that number by 2, getting 1 10 1 10! 1 10 9 8 7 6 = = = 126 2 5 2 5!5! 2 5 4 3 2 1
9 Poker Hands In poker, five cards are dealt from a deck of 52 cards. The types of poker hands (and examples) are 1 Royal Flush: A, K, Q, J, 10 of same suit (10, S, J, Q(, K(, A() 2 Straight Flush: Five cards of same suit in sequence (4(, 5(, 6(, 7(, 8() 3 4 of a Kind: Four cards of the same rank (7(, 7(, 7(, 7() 4 Full House: Three of a kind plus a pair (7, 7, 7, K, K ) 5 Flush: Five cards of the same suit (3, 7, 10, Q, A ) 6 Straight: Five cards in sequence (5, 6, 7, 8, 9 ) 7 3 of a Kind: Three cards of the same rank (J, J, J ) 8 2 Pairs: Two pairs of different rank (5, 5, 9, 9 ) 9 1 Pair: Two cards of the same rank (A, A ) Find a) the total number of poker hands, b) the number of 4-or-a-kinds, c) the number of full houses, d) the number of 3-of-a-kind hands. Solution: a) Total Hands: Each hand represents a subset of size 5 from a set of size 52. Hence the total number of hands a player can receive is 52 choose 5 or 52 52! 52 51 50 49 48 = = = 2,598,960 total hands 5 5!47! 5 4 3 2 13 b) 4-of-a-kind: There are = 13 ways to select the quads and 1 for the remaining card. Using the multiplication principle, we get 48 = 48 1 choices 48 13 = 48 13 = 6244 four of a kind hands 1 1
10 13 12 c) Full House: There are = 13 choices for the rank of the triple, and = 12 1 1 4 choices for the rank of the pair. There are also = 4 ways to choose the triple from a 3 4 card of a given rank, and = 6 ways to choose the pair from four cards of the other 2 rank. Hence, the multiplication principle gives us 13 12 4 4 = 13 12 4 6 = 3,744 full house hands 1 1 3 2 d) Three of a Kind: There are 13 = 13 choices for the rank of the triple, and 1 12 = 66 choices for the rank of the remaining 2 cards. There are also 2 4 4 = 4 choices for the triple of the given rank, and = 4 for each of the 3 1 remaining two cards. Using the multiplication principle, we have 3 13 12 4 = 54,912 3-of-a-kind hands 1 2 1 Naming Results in Mathematics: It is a sorry state of affairs that in mathematics, the most precise of all disciplines, is so lax about giving credit to those who made major discoveries. Pascal s triangle was well known by many mathematicians centuries before Pascal described it in a paper, so why is it called Pascal s triangle? It probably goes back to another person giving credit to Mr. Pascal, and others picking up on that. After a while it becomes Pascal s triangle. Mathematics is rife with all sorts of equations and theorems attributed to one person that were actually discovered by another. Example 7 (Going to the Movies) Three boys and two girls are going to a movie. How many ways can they sit next to each other under the following conditions? Solution a) Neither boy sits next to each other. b) The two girls sit next to each other.
11 a) The only way they can sit is boy-girl-boy-girl-boy. But the boys can be permuted 3! = 6 ways and the girls 2! = 2 ways, so the total number of arrangements is 3!2! = 12. b) First think of the two girls are a single girl so you have 4 persons, 3 boys and 1 girl. Hence there are 4! = 24 ways to permute the two girls among the 3 boys. But, for each of these arrangements, we can permute the two girls 2! = 2 ways, and so the total number of arrangements is 4!2! = 48. Example 8 (Finding Paths) How many paths are there from Start to End in the road system in Figure 5, always moving to the right and down? Solution Counting Paths Figure 5 Solution Since all paths pass through the one-point gap, the problem is subdivided into two parts; finding the paths from Start to the gap, then finding the paths from the gap to End, then multiplying the results together. From Start to the gap, note that we travel a total of eight blocks, four blocks to the right and four blocks down. Labeling each block as R or D, depending whether the move to the right or down, all paths can be written x, x, x, x, x, x, x, x, where four of the x ' s are R and four are D. The total { } number of paths is the number of ways you can select 4 D ' s (or R s) from a
12 set of size 8, which is 8 choose 4 or paths from the gap to End is Combinatorial Card C Trick 3 7 = 35 3 8 = 70. Similarly, the number of 4 Hence, the total number is 8 7 = 70 35 = 2, 450 paths.. 4 3 Here is a little trick you and a friend can perform before an audience. It is very clever and involves two principles of combinatorics, the pigeonhole principle 4 and a set of three elements has 6 permutations. You can be the magician and a friend the assistant. Beginning with a deck of 52 playing cards a volunteer from the audience selects at random five cards from the deck and gives them to the assistant. The assistant looks at the cards, places one card face down on a table, and the other four face up. The magician then makes a grand entrance, looks at the four upright cards, and announces the 5 th card to the audience. How did the magician do it? To show how the trick is done, someone in the audience has selected the cards shown in Figure 6 and gives them to the magician s assistant. Selected Cards Figure 6 3 This trick has been attributed to mathematician William Fitch Cheney. For more variations on this trick, google Cheney s Five Card Trick. 4 The pigeonhole principle is a rather obvious principle from combinatorics which states if m objects are attempted to be placed in n boxes and if m > n, then at least one box contains more than one object. The principle is sometimes stated that if m pigeons try to nest in n pigeonholes and if m > n, then at least one pigeonhole will contain more than one pigeon.
13 For convenience in mental calculations, we the 13 rank values of the cards in a clockwise pattern shown in Figure 7. Counting clockwise no two cards are more than 6 places apart Figure 7 Since there are four suits in the deck and since five cards are selected, the Pigeonhole Principle requires at least one suit will appear more than once. The assistant focuses on this suit (hearts in our example) and then computes the smaller clockwise distance between the 7 and 9, which is 2 (the minimum distance of is always 1,2,3,4,5 or 6). The assistant then places the mystery card, which is the larger 9 face down on a table, and the remaining four cards upright in the order (order is important) 7 2 9 7 The first card of 7 tells the magician the secret card is a heart. The remaining 3 cards 2 9 7 are placed in order which codes one of the numbers 1,2,3,4,5, or 6 and tells the magician how many places past the 7 one must go to reach the mystery card of 9. To accomplish this the magician and assistant agree upon a predetermined ordering of the 52 cards, such as spades first, clubs second, diamonds third, and hearts last and within each suite according to ace low, king high, like A < 2 < 3 < < J < Q < K
14 Hence, the smallest card in the deck is the ace of spades and the largest is the king of hearts. Using this ordering one can code one of the numbers 1,2,3,4,5, S, M, L where or 6 according to one of six permutations of { } A typical code might be S = M = L = smallest rank of the 3 cards middle rank of the 3 cards largest rank of the 3 cards SML = 1 SLM = 2 MSL = 3 MLS = 4 LSM = 5 LMS = 6 In our example the assistant wants to code the number 2 so the 3 cards are displayed in order SLM = 2 7 9. After viewing these cards (and doing some quick mental calculations), the magician decodes the cards and then announces to the surprised audience, the last card to be 9. In case the audience is suspicious that the assistant is sending hidden verbal or visual signals to the magician, another version of the trick might be to have the someone from the audience select five cards from the deck and then give them to the assistant, where after looking at them, the assistant places one of them face down on a table. The assistant then places the other four face up on the table. The magician then makes a grand entrance, looks at the cards for a moment, and then announces the mystery card, then turns it over to a surprised audience. Another variation would be for the assistant to give an audience member the four cards and the audience member say the card to the magician and have the magician read the mind of the audience member to get the last card.
15 Problems 1. Compute the following. a) P (5,3) b) P (4,1) c) P (30,2) d) C (4,1) e) C (10,8) 7 f) 2 9 g) 2 a + b h) 6 a + b i) 10 Permutations and Combinations 2. (Tree Representation of Permutations) Draw a tree diagram showing all permutations of size 2 from the letters abcd. How many permutations are there? 3. (Distinguishable Permutations) The number of distinguishable permuations of 5! the word TOOTS is. Since there are five letters in the word one writes 2!2! 5! in the numerator. However, we cannot distinguish the 2 Ts and the 2 Os in the word, hence we divide 2!2!. Find the number of distinguishable permutations in the following words. a) TO b) TWO c) TOO d) TOOT e) SNOOT f) DALLAS g) TENNESSEE h) MISSISSIPPI i) ILLINOIS 4. (Going to the Movies) Four girls and four boys are going to a movie. How many ways can they be seated if no two girls sit next to each other?
16 5. (Baseball Season) A baseball league consists of 9 teams. How many games will be played over the course of a year if each team plays every other team exactly 20 times? 6. (Hmmmmmmmm) How many 2-element sets are there in the set x N :1 n 100 such that the sum of the 2 elements is even? { } 7. (Picky People) How many ways can 8 people sit next to each other at a movie if a certain 2 of them refuse to sit next to each other? 8. (One Committee) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members? 9. (Two Committees) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members and an entertainment committee of 4 members if no member of the society can serve on both committees? 10. (Three Committees) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members, an entertainment committee of 3 members, and a welcoming committee of 2 members if no member of the society can serve on more than one committees? 11. (Serving on More than One Committee) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members, an entertainment committee of 3 members, and a welcoming committee of 2 members if members can serve on more than one committee? 12. (Counting Softball Teams) A college softball team is taking 25 players on a road trip. The traveling squad consists of 3 catchers, 6 pitchers, 8 infielders, and 6 outfielders. Assuming each player can only play her own position, how many different teams can the coach put on the field? 13. (Permutations as Groups) The 3! 6 illustrated by the 2 3 arrays = permutations of set { } 1,2,3 can be 1 2 3 α = 1 2 3 1 2 3 δ = 1 3 2 1 2 3 β = 2 3 1 1 2 3 ε = 2 1 3 1 2 3 γ = 3 1 2 1 2 3 η = 3 2 1
17 where the bottom row of each array shows how the top row is permuted. Carrying out one permutation followed by another defines a multiplication of the permutations. For instance ε = δβ means we do permutation β first then permutation δ second, which yields the permutation ε (check it yourself). Compute the following. a) αβ b) 2 α c) 3 α d) δεβ After you get the hang of multiplying permutations, make a 6 6 multiplication table of all products. This table describes what in group theory is called the symmetric group of order 3, denoted by S 3. 14. (Catalan Numbers) Catalan numbers represent the number of ways to dissect a polygon into triangles by means of non-intersecting diagonals. Figure xx shows the first 4 Catalan numbers as 1, 2, 5, 14. Can you find the 5 th Catalan number? First 4 Catalan Numbers Figure 6 15.. (Pigeon Hole Principle) Apply the pigeonhole to prove that at least two people in New York City have the same number of hairs on their head. Hint: You may want
18 to make a few assumptions regarding the population of NYC and the maximum number of hairs on the human head. 16. Subsets of a Set. Give another proof that the number of subsets of a set of size n is 2 n. Hint: Assign to each binary number of at most n digits in the following way. Assign a 1 if the corresponding element in the set is selected to be in the subset, otherwise a zero. 17. (Derangements) A derangement is a permutation in which none of the elements 1,2,3 are remain in their natural order. For example the only derangements of ( 3,1, 2 ) and ( 2,3,1 ). Hence we write!3 = 2. Nicolas Bernoulli proved that the number of derangements of a set of size n is n! n = n! k = 1 ( 1) How many derangements are there for the members ( 1, 2,3,4 )? Enumerate them. 19. (Counting Functions) How many functions are there from A { a, b, c} B = { 0,1,2}? Write them down and draw the graphs for a few of them. k! k = to 19. (One-to-One Functions) How many one-to-one functions are there from A a, b, c B = 0,1? = { } to the set of binary numbers { } 20. (Onto Functions) How many onto functions are there from A { a, b, c} set of binary numbers B = { 0,1}? = to the 21. (Counting Functions) Draw graphs of all the functions from A = { a, b, c} to { 0,1 }. Hint: Plot the values a, b, c as points on the x -axis and the numbers 0,1 on the y -axis.