CHAPTER Permutations and combinations

Transcription

1 CHAPTER 3 Permutations and combinations a How many different number plates can be made? b In how many ways can these five people rearrange themselves? c How many different handshakes are possible? d How many different quinellas (first two horses in either order) are possible in a ten-horse race? a C b d

2 78 PERMUTATIONS AND COMBINATIONS In this chapter consideration will be given to certain counting techniques which will enable us to answer questions such as the following: a In how many different ways can I arrange three books, A, Band C, on a shelf? b There are three convenient ways of travelling from Melbourne to Sydney -namely, by aeroplane, by bus or by car. In how many different ways can I travel from Melbourne to Sydney and back to Melbourne again? c Car registration plates in Victoria consist of three letters of the alphabet followed by three numerals. How many different number plates can be made? d In how many different ways can five girls, A, B, C, D and Ebe arranged in a row, taken: (i) three at a time? (ii) five at a time? e A, B, C, D and E are five different points on the circumference of a circle. How many chords can be drawn between these points? f In how many different ways can six numbers be drawn out of 45 in a Tattslotto draw? g How many different hands of five cards can be dealt from a pack of 52 playing cards? 3.1 Permutations Example 1 In how many different ways can I arrange three books, A, Band C, on a shelf? We could list the possible arrangements and then count them: ABC,ACB,BAC,BCA, CAB, CBA There are six different arrangements. The information could be set out neatly with the aid of a tree diagram: First space Second space Third space C A B C B A : B C A Arrangements ABC ACB BAC BCA CAB CBA Figure 3-1 We could also consider the situation with the aid of boxes. 3 2 Figure 3-2 The first space can be filled in three ways, because any one of the three books can occupy the first space. When the first space is filled in any one of these three ways, the second space can be filled in two ways and, since each way of filling the first space can be associated with each way of filling the second space, the total number of ways of filling the first two spaces is 3 x 2, i.e. six ways. When the first two spaces have been filled in any one of these six ways, there is only one book left to fill the third space. So the number of ways of filling the three spaces is 3 x 2 x 1, i.e. six, ways.

3 PERMUTATIONS AND COMBINATIONS 79 It is important to note that ABC and A CB, for example, are two different arrangements. Each different arrangement is called a permutation. Order must be taken into account. Our reasoning in this case involved the use of the multiplication principle. 3.2 The multiplication principle If one operation can be performed in m different ways and if, when it has been performed in any one of these ways, a second operation can be performed in n different ways, the number of ways of performing the two operations is m x n. Example 2 There are three convenient ways of travelling from Melbourne to Sydney - namely, by aeroplane (A), by bus (B) or by car (C). In how many different ways can I travel from Melbourne to Sydney and back to Melbourne again? Suppose I travel to Sydney by aeroplane (A). I can then return to Melbourne in three different ways - namely by A, B or C. Similarly, if I travel to Sydney by bus (B), I can return to Melbourne in three different ways - namely by A, B or C. Likewise, if I travel to Sydney by car (C), I can return to Melbourne in three different ways - namely by A, B or C. Since each of the three ways of travelling from Melbourne to Sydney can be associated with each of the three ways of returning to Melbourne, there are 3 x 3, i.e. nine, different ways of travelling from Melbourne to Sydney and to Melbourne again (multiplication principle). We can also consider the situation with the aid of boxes. Figure The first box can be filled in three ways since there are three ways of travelling from Melbourne to Sydney. When the first box is filled in any one of these three ways, the second box can then be filled in three ways and, since each way of travelling from Melbourne to Sydney can be associated with each way of travelling from Sydney to Melbourne, the total number of ways of travelling from Melbourne to Sydney to Melbourne is 3 x 3, i.e. nine, ways. The situation can also be neatly represented by the following tree diagram. Figure 3-4 Melbourne to Sydney to Possible Ways Sydney Melbourne AA A B C AB AC BA BB BC CA CB cc

4 80 PERMUTATIONS AND COMBINATIONS The multiplication principle can be extended to more than two operations. If one operation can be performed in m different ways and if, when it has been performed in any one of these ways, a second operation can then be performed in n ways and, after that, a third operation can be performed inp different ways, etc., then the successive operations can be performed in mnp... ways. Example3 Car registration plates in Victoria consist of three letters of the alphabet followed by three numerals. How many different number plates can be made? Figure 3-5 This is equivalent to filling six places, the first three with letters of the alphabet and the last three with the numerals Oto 9. The first place can be filled in 26 ways, since any one of the 26 letters of the alphabet can be used. When this place has been filled in any one of these 26 ways, the second place can also be filled in 26 ways, since the same letter can be used in both places. Applying the multiplication principle, the first two places can be filled in 26 x 26 different ways. It follows that, extending the multiplication principle the total number of plates that can be made.is: 26 X 26 X 26 X 10 X 10 X 10, i.e This is a rather large number! It would be rather messy to represent this situation with a tree diagram. Example4 In how many different ways can five girls, A, B, C, D and Ebe arranged in a row, taken three at a time? Figure 3-6 The first place can be filled in five ways, since any one of the five girls can occupy this place. When it has been filled in any one of these five ways, the second place can then be filled in four ways and the third place in three ways and, by application of the multiplication principle, the total number of arrangements is: 5 X 4 X 3, i.e. 60. This question could be put in many different contexts: a In how many different ways can we appoint a captain, vice-captain and committee member from the five girls? b In how many different orders could these five girls fill first, second and third places in a 100-metre race? Can you think of other similar contexts?

5 \ PERMUTATIONS AND COMBINATIONS Mutually exclusive operations: addition principle Example 5 How many two-digit or three-digit numbers can be formed from the digits 1, 2, 3,... 8 if the same digit may not be used more than once in any number? a Two digit numbers: 8 7 Figure 3-7: Total 56 b Three digit numbers: Figure 3-8: Total 336 There are 5i = 392 numbers that contain two digits or three digits. a and b are said to be mutually exclusive or disjoint operations, i.e. the formation of a twodigit number excludes the formation of a three-digit number; you may have one or the other kind of number but not both at the same time. If two operations are mutually exclusive, and the first operation can be performed in m different ways and the second in n di ferent ways, then one operation or the other can be perf armed in m + n different ways. 3.4 Definition of permutation Each of the ordered subsets that can be formed by selecting some or all of the elements of a set is called an arrangement or permutation. The verb 'to permute' means 'to interchange'. Verify that Examples 1 to 5 above conform to this definition. Exercises 3a 1 There are four roads from town A to town B,-and three from town B to town C. How many different ways are there of travelling from A to B to C? 2 A man has three pairs of shoes, four suits and six ties. How many different sets of shoes, suits and ties can he wear? 3 A restaurant menu provides three choices of soup, four choices of main course and three choices of sweets. How many different meals of soup, main course and sweets are available? s 4 Anna, Lubica, Guiseppe and David are members of a tennis team from which a cap,tain and a vice-captain are to be elected. In how many different ways can this be done? I (

6 82 PERMUTATIONS AND COMBINATIONS r J 5 a In how many ways can a first, second and third prize be awarded in a class of 10 students? b In how many ways can the Mathematics, Physics and Chemistry prizes be awarded in a class of 10 students? 6 In how many ways can the positions of president, vice-president, secretary and'treasurer be filled from a committee of six people? 7 In a certain State, number plates for cars consist of two letters of the alphabet followed by two of the digits Oto 9. a How many different number plates can be made? b How many of these begin with the letter A? I 8 There are four roads from town A to town B, and three from town B to town C. Three other roads by-pass B, and go direct from A to C. How many different ways are there of travelling from A to C and back to A? 1 9 Four different airline companies and three different shipping companies operate between Sydney and Auckland. How many different ways may a person travel: a from Sydney to Auckland? b from Auckland to Sydney? c from Sydney to Auckland and back to Sydney? 10 In how many ways can the letters of the word 'pencil' be arranged? " 11 In how many orders can five runners pass the winning post, if no two pass it together? 12 In how many ways can first, second and third place be allotted to five runners if no two pass the winning post together? 13 Using five flags of different colours, how many different signals can be made by hoisting them on a vertical mast, if at least three flags are used for each signal? \ 14 How many different people can there be, if each person has two initials, but no two have the same ordered pair of initials? 15 a In how many ways can three letters be posted in five different letter boxes? b In how many ways can this be done if each letter must be posted in a different box? 16 A man has three pairs of shoes (one of which is coloured brown), four suits (one of which is coloured blue) and six ties. a How many different sets of shoes, suits and ties can he wear? b How many combinations are possible if he does nqt wear brown shoes with his blue suit? 17 How many different arrangements of three letters of the word 'Sunday' are possible? 18 From the digits 4, 5, 6, 7, find: a how many two-digit or three-digit numbers can be formed b how many numbers containing at least two digits can be formed c how many different numbers can be formed. Assume that no digit can be used more than once in a number. 19 A ship sends signals by hoisting four different flags on a vertical mast. a How many different signals can be sent? (r How many different signals can be sent by hoisting at least one of the four flags?,,.., I Ir 20 There are four different roads from town A to town B, three from town B to town C, five from town A to town D and three from town D to town C. How many different ways are there of going from A to C via B or D?

7 PERMUTATIONS AND COMBINATIONS In how many ways can three letters be mailed in five letter boxes if there is no restriction on which boxes are to be used? 22 a How many 4-digit numbers can be formed? b How many of these end in 2, 4 or 6? 23 Polygons are usually labelled by placing letters of the alphabet at their vertices. How many different ways are there of labelling: a a triangle? b a pentagon? 24 The twenty players in a football team elect a captain and vice -captain from their ranks. In how many ways is this possible? 3.5 The symbol np,. In Example 4, we s_aw that the number of arrangements of the five girls, A, B, C, D and E, taken three at a time, was 5 x 4 x 3, i.e. 60. The symbol 5 P3 denotes the number of arrangements of five objects taken three at a time. i.e. 5 P 3 = = 60 The symbol 6 P 2 denotes the number of arrangements of six objects taken two at a time. i.e. 6 P2 = 6. 5 = 30 The symbol 6 P6 denotes the number of arrangements of six objects taken six at a time. i.e. 6 P 6 = = 720 The symbol 6 P6 can be written in a concise form; it is the product of six consecutive integers from 6 to 1 and is written as 6! (factorial 6). The product of n consecutive positive integers from n to 1 is denoted by n! and is called factorial n or n factorial. I n! = n(n - l)(n - 2)... 3 x 2 x 1 I For example, the symbol 5 P 3 above can be expressed in factorial form: 5 P3 = ! 2! Similarly: 6 P2 = 6. 5 = ! 4! 6! = 4!

8 84 PERMUTATIONS AND COMBINATIONS The symbol n p r denotes the number of arrangements of n different objects taken rat a time, i.e. the number of ordered r-subsets that can be formed from an n-set. It denotes the number of ways of filling r places with n different objects at our disposal. (r n). n n-1/ n-2 n - r+ 1 Figure 3-9 The first place can be filled in n ways, because any one of the n objects can occupy this place. When the first place has been filled in any one of these ways, there remain (n - 1) objects, any one of which can occupy the second place. Since, by the multiplication principle, each way of filling the first place can be associated with each way of filling the second place, the number of ways of filling the first two places is n(n - 1). Similarly, when the first two places have been filled in any one of these ways, there remain (n - 2) objects, any one of which can occupy the third place. So the first three places can be filled in n(n - l)(n - 2) ways, and so on. It follows that the number of ways of filling r places is given by: n pr = n(n - l)(n - 2)... (n - r + 1) _ n(n - l)(n - 2)...(n - r + l)(n - r)! - (n - r)! =--- n! (n - r)!... (1)... (2) The symbol n p n then, would denote the number of ways of filling n places with n objects at our disposal. But: So: n P n = n(n - 1 )(n - 2) = n! from (1) n p n! n! n = ( _ )I = from (2) n n. n '. = n! O! For this to be true, we define O! to be 1. Computer applications 10 REM Program to evaluate n! 20 READ N 30 F = 1 40 C = 1 50 F = F*C 60 C = C IF C < = N THEN PRINT N; "FACTORIAL EQUALS"; F 90 DATA END Explain how the lines in the above program provide for the automatic calculation of n! Run the program for different values of n by amending line 90.

9 PERMUTATIONS AND COMBINATIONS 85 1 O REM Program to evaluate npr tor 1 i;;; r i;;; n 20 READ R, N 30 P = N 40 FOR C = 1 TO R P = P*(N - C) 60 NEXTC 70 PRINT "VALUE OF"; N; "P"; R;" = "; P 80 DATA3, 7 90 END Explain how the lines in the above program provide for the automatic calculation of np,, Run the program for different values of n and r by amending line Arrangements with restrictions Example 6 In how many ways can six girls and two boys be arranged in a row if: a the two boys are together? b the two boys are not together? a Since the boys are to be together, they may be regarded as one unit. So there are seven objects (the six girls and the unit of two boys) to be arranged in a row. This can be done in 7! ways. However, the two boys, whom we shall denote by A and B, can be arranged among themselves in 2! (or 2) ways, namely AB or BA. Number of arrangements = 2! 7! = = b Number of arrangements without restriction = 8! Number of arrangements with the boys together = 2. 7! Number of arrangements with the boys not together = 8! - 2.7! = 8. 7! ! = 6. 7! = Example 7 How many different arrangements of the letters of the word 'Tuesday' are possible if: a the three vowels are to be together? b the first and last places are to be filled by consonants? c the consonants and vowels are to occupy alternate positions? a Since the three vowels are to be together, regard them as one unit. So there are five objects (the other four letters and this unit consisting of u, e and a) to be arranged in a row. This can be done in 5! ways. However, the vowels remaining together can be arranged among themselves in 3! ways. Number of arrangements = 3! 5! = = 720 \

10 86 PERMUTATIONS AND COMBINATIONS b There are restrictions on the first and last places. It is simplest then to fill them first Figure 3-1 O Since there are four consonants, the first place can be filled in four ways. When the first place is filled in any one of these four ways, the last place can be filled in three ways. So there are 12 ways of filling the first and last places. When these places have been filled in any one of these 12 ways, there are five places left which can be filled without restriction in 5! ways. Number of arrangements = 12.5! = 1440 c Since the consonants and vowels are to occupy alternate positions, the pattern would have to be as shown below. C V C I V C V C Figure 3-11 Consonants occupy the first, third, fifth and seventh places, and vowels occupy the second, fourth and sixth places. So the consonants can be arra]j.ged in 4! ways and the vowels in 3! ways. Number of arrangements = 4! 3! = 144 Exercises 3b 1 Evaluate: a 5 p 3 b 4 p2 C 6 p5 d B p 4 e 9 pl f l0 p4 g "P2 h "P 4 2 Evaluate: a 5! b 4! C 6! d 8! e 5! f 7! 10! 12! 2! 6! g h 8! 9! 3 Write the symbol which denotes the number of: a permutations of eight different objects taken three at a time. b ways of arranging five objects when we havy seven objects at our disposal. c ordered 4-subsets that can be formed from a 6-set. Evaluate in each case. 4 a How many five-digit numbers can be formed from the digits 3, 4, 5, 6 and 7, if no digit can be used more than once in a number? b How many even five-digit numbers can be formed? c How many of the five-digit numbers in Question 4a are divisible by 5?

11 PERMUTATIONS AND COMBINATIONS 87 5 a In how many ways can three men and three women be arranged in a row so that the men are always together? b In how many ways can they be arranged without restriction? c In how many ways can they be arranged so that the men and women occupy alternate positions? 6 a How many different arrangements of the letters of the word 'iiµic10 are possible? b How maf!.y such arrangements have: (i) the vowels together? (ii) the vowels occupying the end positions? (iii) the vowels and consonants occupying alternate positions? 7 An athletics meeting consists of five sprints and three hurdles. In how many different ways can the events be arranged so that the meeting starts and finishes with a hurdle race? 8 A train has seven carriages, three of which contain sleeping compartments. In how many ways can the carriages be arranged so that those containing sleeping compartments are at the rear? 9 In how many different ways can four boys and three girls be arranged in a row so that: a the boys and girls occupy alternate positions? cy the end positions are occupied by boys? c the girls are together? d two particular boys are together? 10 a In how many ways can the letters of the word 'plants' be arranged in a row? b How many of these arrangements begin with p and end with s? 11 How many even numbers of four digits each can be formed with the digits 3, 4, 7 and 8: a if no digit is repeated? b if repetitions are allowed? 12 a In how many ways can three different Mathematics books, three different Physics books and two different Chemistry books b.e arranged-on a shelf? b In how many of these arrangements are the Mathematics'books together? 13 How many fiy!=!-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if the same digit may be used more than once in a number, and if each number is: a divisible by 5? b an even number? c an odd number? d divisible by 4? 14 In a group of five women and four men, there is one married couple. In how many different ways can the nine people arrange themselves in a row if: a the married couple is together? b the married couple is not together? c the men and women occupy alternate positions? d the women are together? 15 Find t1ie number of different ways in which n children, two of whom are boys and the rest girls, can stand in a row, if the two boys stand together. 16 A chessboard has 64 squares arranged in eight rows of eight. The only moves that a King can make are from the square on which it is standing to an adjacent square. Find the total number of moves, from all possible positions, that the King can make.

12 88 PERMUTATIONS AND COMBINATIONS 17 How many five-letter words, i.e. arrangements of five letters (pronouncable or not), can be made in which vowels and consonants alternate - no letter being used twice. I.._ 18 Three boys and two girls sit side-by-side on five chairs arranged in a row. In how many ways can they arrange themselves: a if the two girls do not sit next to each other? b if the end positions are occupied by boys? 3. 7 Arrangements in a circle Example 8 In how many ways can four women, A, B, C and D, be arranged in a circle? There are two ways of reasoning: (i) Since there is no first place to fill we can take one of the women and put her in a fixed position and arrange the other three around her. This can be done in 3! ways. A D B Figure 3-12 C or (ii) Figure 3-12 shows one arrangement of A, B, C, and D. If we keep the four women in the same position relative to one another and move them one place clockwise, we still have the same arrangement. On the other hand, if they were arranged in a straight line and we moved them one position to the right so that the woman in the fourth position then occupied the first position, this would be a different arrangement (Figure 3-13). IAIBICl 0I Figure 3-13 I l 0IAIBICI ICIDIAIBI 1 cl 01 I 8 A In other words, for every one arrangement in a circle, there would be four different arrangements in a straight line. Number of circular arrangements = ; = 3! =6 The number of ways of arranging n different objects in a circle, regarding clockwise and anticlockwise arrangements as different, is: '\ I ' i.e. (n - 1)!

13 PERMUTATIONS AND COMBINATIONS Number of arrangements of n objects in a row, when they are not all different Example9 Seven tiles are identical except in colour. Three of the tiles are red, two are white and two are black. The tiles are arranged in a row. How many different arrangements are po sible? Since there are three red (R), two white (W) and two black (B) tiles: 7! the number of arrangements = 3! = 210 Reason: One of the 210 arrangements is RRRWWBB. If the three red tiles were different colours - green, yellow and orange for example - this arrangement would be: GYOWWBB or YGOWWBB or OGYWWBB or or or GOYWWBB YOGWWBB OYGWWBB These are 3!, i.e. 6, different arrangements. This accounts for the division of 7! by 3! Reasoning similarly with the two white and the two black tiles, the number of. 71 arrangements 1s 3! The number of ways of arranging n objects in a row, when there are p alike of one kind, q alike of another, etc. is: n! p!q!... Exercises 3c 1 a In how many ways can eight people be seated at a round table? b Would the number of arrangements be the same if the table were: (i) rectangular? (ii) semicircular? 2 Two women and five men are to be seated at a round table. How many different arrangements are possible if the two women are to sit together? 3 In how many ways can the six letters of the word 'mammal' be arranged in a row? 4 Seven identical cubes, folir of which are red and thiee of which are black, are arranged in a row. How many different arrangements are possible? 5 Three red, three white and three blue balls are placed in a row'. a How many distinguishable arrangements are possible? b In how many of these arrangements are the red balls together? 6 How many seven-digit numbers can be formed, each containing all the digits 2, 3, 3, 3,4,5and6? 7 How many numbers greater than 6000 can be formed with the digits 3, 4, 5, 6 and 7 if a digit cannot occur more than once in a number?,!

14 90 PERMUTATIONS AND COMBINATIONS,, 8 In how many ways can six people be arranged in a circle if two particular people are always: a together? b separated? ), 9 Father, mother and six children stand in a ring. In how many ways can they be arranged if father and mother are not to stand together? 1 0 If the women must stand together, in how many ways can five boys and three women be arranged: a in a row? b in a circle? 4 11 Four men and four women are to be seated alternately: a in a row, b at a round table. In how many ways can this be done? 12 How many even numbers of four digits each can be formed with the digits 3, J 7, and,j; J a if no figure is repeated? b if repetitions are allowed? 13 How many numbers greater than 4000 can be formed from the digits 3, 5, 7, 8 and 9? (Repetitions are not allowed.) 14 In how many ways can the letters of the word "penfud ' be arranged if:, a the first and last places are occupied by consonants? b the vowels and consonants occupy alternate places? 15 Find the number of different arrangements which can be made from seven identical. blocks whose colours are violet, indigo, blue, green, yellow, orange and red, such that the blue and the green are not to be together and the blocks are to form the following patterns: a b I' C ' r \ 16 How many five-letter code words can be made in which vowels and consonants alternate and no two letters are used twice? 17 Find t,he number of arrangements of the letters in the word 'pencils' if: a 'e' is next to 'i' (_ 'e' precedes 'i' c there are three letters between 'e' and 'i'. 18 a In how many ways can the letters of the word ririciple' be arranged?. b In what proportion of these arrangements do the letters 'p' come together? 19 a In how many ways can the letters in the word 'pr 'ciy;j#n' be arranged? b In how many of these arrangements do the vowels occupy even places?

15 PERMUTATIONS AND COMBINATIONS 91 ' 20 How many arrangements can be made by the letters of the word 'definition': a if the letters 'i' do not occupy either the first or the last place? b if the letters 'i' are together? 21 How many arrangements of the letters in the word 'tomato' are there, if the letters 'o' are to be separated?. I. \ ' J ' 22 A car can hold three people in the front seat and four people in the back seat. In how. many ways can five people be seated in the car if two particular people must sit in the bacf seat and one particular person is the driver? 23 In how many ways can four people be accommodated in a certain hotel if there are four rooms available? 24 In how many ways can eight oarsmen be seated in an eight-oared boat if three can row only on the stroke side and three can row only on the bow side? 25 Prove (i) from the definition of n P, and (ii) from the formula for n P,, that: n+i p, = n p, + r. n p, _ 1 26 In how many ways can five girls and five boys be arranged in a circle so that the girls are separated? In how many ways can this be done if two particular boys must not be next to a particular girl? /,!.' 27 How many arrangements of the letters of the word 'ParfarfulLta:' are possible? 28 The diagram below shows the pattern of roads in a housing estate. In how many different ways can you go from A to B? You must not go along the same stretch of road twice, although you may pass the same corner a second time. You are not permitted, at any stage after leaving A, to travel back towards the centre...,. 2 The diagram below shows a pattern of suburban roads. In how many different ways can a motorist travel from A to X without turning back at any stage? 'D'D'CJ :o:o:o:.0,01d, )JOD1D, 30 Each of a set of equal cubes has its six faces painted red, orange, yellow, green, blu 1 e and white respectively so that no two cubes are alike and every arrangement of colours is used. How many cubes are there in the set?

16 92 PERMUTATIONS AND COMBINATIONS 3.9 Combinations Each of the subsets that can be formed by selecting some or all of the elements of a set without regard to the order in which the elements appear in the subset is called a selection or combination. The distinction between a permutation and a combination can be illustrated by a simple example. Consider a set of four people, whom we shall denote by A, B, C and D. In how many ways can they be arranged, taking them two at a time? How many selections of two can be made? To solve this problem, we can list the possible arrangements and selections and count them. The possible arrangements are: AB, AC, AD, BC, BD, CD, BA, CA, DA, CB, DB, DC. There are 12 arrangements. The number of selections is only six, namely: AB, AC, AD, BC, BD, CD. Note that AB and BA are two different arrangements but only one selection of two people. Order must be taken into account in permutations. In combinations, however, we are concerned only with the number of elements in each subset. Each of these groups contains two people, who can be arranged among themselves in 2! ways. So, number of combinations x 2! = number of permutations. If we denote the number of combinations of four people taken two at a time by the symbol (1) or 4 C2, then: ( i ) 4p =-2 2! Note: The symbol (1) is read '4 above 2'. Example 10 How many groups of three can be selected from four girls, A, B, C and D? The symbol (;) or 4 C3 denotes the number of 3-subsets that can be formed from a 4-set. The subsets in this example can be listed by inspection; ABC, ABD, ACD, BCD. Consider the subset ABC. These three girls can be arranged among themselves in 3! ways to give six different arrangements: ABC,ACB,BCA,BAC, CBA, CAB.

17 PERMUTATIONS AND COMBINATIONS 93 Similarly, the other three subsets can be arranged in six ways as shown in this table: Combinations ABC ABD ACD BCD Permutations ABC,ACB,BCA,BAC,CBA,CAB ABD,ADB,BDA,BAD,DBA,DAB ACD,ADC,CDA,CAD,DCA,DAC BCD,BDC,CDB,CBD,DCB,DBC So, number of combinations x 3! = number of permutations. That is: (1) X 3! = 4P3 and so: = 4 p3 3! = = The symbol( ) or - This symbol denotes the number of combi s of n different objects taken rat a time, i.e. the number of,-subsets that can be formed from an n-set. Each of these combinations consists of a group of r different elements that can be arranged among themselves in r! ways. So (;) x r! = number of arrangements of n different objects taken rat a time = np,, _ n(n - l)(n - 2)... (n - r + 1) - r! This formula for (;) may also be written in factorial form - if we multiply numerator and denominator by (n - r)!, we get: ( n r ) = n(n - l)(n - 2)...(n - r + l)(n - r)! r!(n - r)! n! =---- r!(n - r)! where (;) is read 'n above r'. For example: ( ! 3 ) = or 3! 3!4! = 35 = or _fil_ 5! 5!3! = 56

18 94 PERMUTATIONS AND COMBINATIONS The number of combinations of n objects taken rat a time is equal to the number of combinations of n objects taken (n - r) at a time; that is: For each set of r objects selected, there is left behind a set containing (n - r) objects, and so there are as many sets containing (n - r) objects as there are sets containing r objects. Alternatively: n! (n r) (n - r)!(n - n - r)! n! (n - r)!r! = (;) If, from a set of nine boys, we select a set of six, a set containing three boys is left behind. G) = G) = = If, from a pack of 52 playing cards, we select a hand of 50 cards, we automatically also select a hand of 2 cards. G ) = (5;) = 1326 Computer applications 1 0 REM Program to evaluate ncr 20 READ R, N 30 P = N 40 FOR C = 1 TO R P = P*(N - C) 60 NEXT C 70 GOSUB S = P/F 90 PRINT "VALUEOF"; N; "C"; R;" ="; S 100 DAT A4, END 200 REM Subroutine for r! 210 F = FOR K = 1 to R 230 F = F*K 240 NEXT K 250 RETURN

19 PERMUTATIONS AND COMBINATIONS 95 Explain how the lines in the above program provide for the automatic calculation of "C, i.e. (;). Run the program for different values of n and r by amending line Example 11 A, B, C, D and E are five points on the circumference of a circle. How many different chords can be drawn between these points? (See Example 1 of Chapter 7 in Change and Approximation.) B A chord is a straight line joining two points on the circumference of a circle. So, if we have five points at our disposal, we are asked to find the number of ways of s lecting the five points two at a time. This can be done in (D ways. (D = 10. There are 10 chords. Represent this situation with a tree diagram. Figure 3-14 ; r r- Example 12 A group of 20 students consists of 12 girls and eight boys. A committee of five is to be selected to make arrangements for an excursion. a How many committees of five can be formed? b How many committees contain three girls and two boys? c How many committees contain at least three girls? a Number of committees = ( 2 5 ) = b Three girls can be selected from 12 in 12 ( 3 ) ways. Two boys can be selected from eight in (D ways. Number of committees = C32 ) (D 'Ll = 6160 ( 1 2 is 3 ) multiplied by (D because the multiplication principle applies.

20 96 PERMUTATIONS AND COMBINATIONS Note: c The committees may consist of: (i) three girls and two boys (ii) four girls and one boy (iii) five girls and no boys Number of possible committees = (1}) (D + (1;) (D + C5 2 ) ( ) = LJ '1 v Addition in this case because the operations are mutually exclusive. ( ) ( ) ( n ) 8!. = 0! 8! usmg r = 1, since 0! n! = 1 =-o,. n. ' - n! - r! (n - r)! Example 13 In TAB betting, the 'trifecta' pays on the first three horses in correct order; and the quinella pays on the first two horses in either order. In a 12-horse race, what is the possible number of: a trifecta combinations? b quinella combinations? a Since the order in which the first three horses finish is important in trifecta combinations, we need to find the number of permutations of 12 horses taken three at a time. Figure Number of trifecta combinations = 12 P 3 = = 1320 b Since the order in which the first two horses finish is not important in quinella combinations, we need to find the number of combinations of 12 horses taken two at a time. Number of quinella combinations = ( 1 }) = z:-t = 66 Example 14 In a maths test consisting of 10 questions, students are required to attempt seven questions. How many choices are available to the students if the first question is compulsory and the last question carries a bonus and is not to be included in the seven questions?

21 PERMUTATIONS AND COMBINATIONS 97 Students are required to select six questions out of the eight questions numbered 2to 9. Number of choices Example 15 In how many ways can six people be separated into three sets containing 3, 2 and 1 respectively? Three people can be selected from six in ( ) ways. From the remaining three people, two can be selected in (D ways. This leaves one person for the remaining set. Number of ways = ( ). G) = 6.5.4_ = 60 Exercises 3d 1 Evaluate: a (D b G) C (!) d (;) e 12 C 4 2 Use the fact that ( ; ) = (n r) to evaluate: f tsco g 6 Cs h n c s a (18 0 ) b G ) C GD d ( ) e ( ) f 4 C 36 g 48 C 4s h 2 C1s 3 Write the symbol that denotes each of the following: a The number of 3-subsets there are in a 6-set. _ b The number of combinations of n different things taken two at a time.,' c The number of ways of selecting six objects when we have 15 to select from. d The number of ways a tennis team of four players can be selected from 12 players. e The number of ways a cricket team of 11 players can be selected from 14 players. f The number of ways a football team of 18 players can be selected from 20 players. g The number of hands of five cards that can be dealt from a pack of 52 playing cards. h The number of ways of drawing six numbers out of 45 in a Tattslotto draw. i The number of ways a jury of 12 can be selected from 15 people. Evaluate in each case. 4 A, B, C, D, E and Fare six different points on the circumference of a circle. a How many triangles can be formed by joining the points? b How many triangles can be formed if there are: (i) 10 points? (ii) 15 points? (iii) n points? ( ( () ' ' <

22 98 PERMUTATIONS AND COMBINATIONS 5 Eight people are introduced and shake hands with each other. a How many handshakes are there? b How many handshakes are there if there are: (i) 10 people? (ii) 20 people? (iii) n people? 6 How many triangles can be formed by joining the vertices of: a a convex pentagon (five sides)? b a convex octagon (eight sides)? c a convex dodecagon (12 sides)? 7 Along a railway line there are 20 stations, both terminals included. How many different tickets (one class only and single) would be required to connect each station with every other station? '1 ; 8 The eleven players of a cricket team elect from their ranks a captain and vice-captain. In how many different ways is this possible? 9 a How many different hands of five cards can be dealt from a pack of 52 playing cards? b How many of these contain the ace of hearts? 10 a In how many ways can a committee of three men and four women be chosen from six men and seven women? Wha,proportion of these committees contain a particular man and a particular woman? 11 Fifteen different subjects are available to some VCE students. They have to select six subjects, two particular subjects of the 15 being compulsory. How many different choices are available? 12 a How many different committees of four can be formed from a group of eight students? b How many of these contain one particular student, X? c How many contain two particular students, X and Y? d How many do not contain X? f $ / /(1-r 13 In how many different ways can a group of: a three boys be selected from eight boys? b two girls be selected from five girls? c three boys and two girls be selected from eight boys and five girls? 14 In how many different ways can a selection of five be made from 12 books if one particular book is to be included in the five? In how many different ways can a set of two boys and three girls be selected from five boys and four girls? 16 In how many ways can a committee of four men and five boys be formed from eight men and seven boys? 17 In how many ways can a jury of 12 be chosen from 10 men and 7 women if it contains: a eight men? b at least eight men? 18 A cricket eleven is to be selected from 15 players ( eight are batsmen and seven are bowlers). How many teams are possible if the team must contain: a at least six batsmen? ', b no fewer than four bowlers and no more than six bowlers? )

23 PERMUTATIONS AND COMBINATIONS Ho\. many selections of five cards can be made from a pack of 52 playing cards so that there are: a at least three aces? b three hearts? c at least one heart? d not more than two spades? 20 From a class of IO girls and IO boys, how many different committees of five can be selected that contain: a exactly three girls? b not more than three girls? c at least one boy and at least one girl? 21 How many selections of five pieces of fruit can be made from four oranges, three bananas and two apples if a selection must contain: a exactly three oranges? b exactly one apple and at least' two bananas? 22 h how many ways can three cards be selected from a pack of 52 playing cards if: a at least one of them is an ace? b not more than one is an ace? 23 A carton contains 10 cubes, six of which are black and four of which are white. A random sample of five cubes is selected. How many different selections are possible if each sample contains: a exactly four black cubes? b at least four black cubes? c no more than four black cub s? '<',f 24 From seven teachers and five pupils a committee of seven is to be formed. How many committees can be selected if both teachers and pupils are represented, and the teachers are in a majority? 25 From four oranges, three bananas and two apples, how many selections of five pieces of fruit can be made, taking at least one of each kind? 26 In how many ways can a jury of 12 be chosen from 10 men and seven women so that there are at least six men and not more than four women on each jury? 27 How many (i) selections and (ii) arrangements consisting of three consonants and two vowels can be made from eight consonants and four vowels? /[, - 28 a In how many ways can four Physics books and three Mathematics books be arranged on a shelf if a selection is rhade from six different Physics books and five different Mathematics books? b In how many of these arrangements are the Physics books together? 29 If G) = (;), find the value of n. 30 The ratio of the number of combinations of (2n + 2) different objects taken n at a time to the number of combinations of (2n - 2) different objects taken n at a time is 99:7. Find the value of n. 31 In how many ways can nine books be distributed among a man, a woman and a child, if the man receives four, the woman three and the child two? > 1 32 In how many ways can eight boys be divided into two unequal sets? b,_

24 100 PERMUTATIONS AND COMBINATIONS 33 In how many ways can three boys and two girls be arranged in a row if a selection is made from five boys and four girls? In how many of these arrangements does a boy occupy the middle position? 34 How many words (arrangements of letters), containing three consonants and two vowels, can be formed from the letters of the word 'promise'? 3.11 Probability associated with permutations and combinations Permutations On the assumption that, if n objects are arranged in a row (or a circle), each arrangement is equally likely and, if there are certain restrictions imposed which reduce the number of possible arrangements, then we may consider probability as simply a ratio of the number of favourable arrangements to the number of possible arrangements. Pr = number of favourable arrangements number of possible arrangements Example 16 Six girls and two boys arrange themselves at random in a row. What is the probability that: a the two boys are together? b the two boys are not together? (See Example 6.) a A sample space of the experiment of arranging eight people in a row may be considered as consisting of 8! points, each equally weighted. Number of possible arrangements = 8! Since the boys are to be together, they may be regarded as one unit. So there are seven objects (the six girls and the unit of two boys) to be arranged in a row. This can be done in 7! ways. However, the two boys, whom we shall denote by A and B, can be arranged among themselves in 2! (or: 2) ways, namely AB or BA. Number of favourable arrangements = 2! 7! Pr(the two boys are together) = 8! 2. 7! 8. 7! 1 4 b Number of arrangements without restriction = 8! Number of arrangements with the boys together = 2. 7! Number of arrangements with the boys not together = 8! ! = 8. 7! ! = 6. 7! 6. 7! Pr(the two boys are not together) =8! _ 6. 7! ! = 3 4 Compare the answers for a and b. What do you observe?

25 PERMUTATIONS AND COMBINATIONS 101 Example 17 > ) v,,, l Seven tiles are identical except for colour. Three of the tiles are red, two are white and two are black. The tiles are arranged at random in a row. What is the probability that the red tiles are together? 7 ' (See Example 9.) 71. N um b er o f arrangements wit h out restnctlon = 3! Since the red tiles are to be together, consider them as one unit. So we have five units to arrange in a row, two of which are white and two are black. The red tiles, remaining together, can be arranged among themselves in only one way. 51 Number of favourable arrangements = Pr(the red tiles are together ) = X 7i 1 7 Example 18 A student is asked to write any three-digit number using the digits 1 to 9. What is the probability that the three digits are the same? Number of possible numbers = 9 x 9 x 9 There are only nine favourable numbers 111, 222, Required probability = 9 x x Combinations [sampling without replacement] Example 19 In TAB betting, the 'quinella' pays on the first two horses in either order. In a 12-horse race, what is the probability of picking a quinella, assuming that each horse is equally likely to win. The number of quinella combinations is equal to the number of ways of selecting two horses when we have 12 to select from. Number of quinella combinations = ( 1,J) = 66 Only one of these combinations is favourable. 1 Pr= 66 Problems associated with selections frequently deal with the taking of a sample without replacement from a small population consisting of two or more different kinds of i ems. On the assumption that all selections are equally likely we may consider probability as a ratio of the number of favourable selections to the number of possible selections. Pr = number of favourable selections number of possible selections \

26 102 PERMUTATIONS AND COMBINATIONS Example20 An urn contains nine distinguishable cubes of which three are white and six are black. Two cubes are drawn at random without replacement. Calculate the probability that both cubes are black. A sample space of the experiment of drawing two cubes without replacement from an urn containing nine cubes may be considered as consisting of ( ) points, each equally weighted, ( ) points corresponding to the event 'both cubes are black'. So: G) Pr(both cubes are black) = ( ) Ll'9.8 = 12 Alternative method Let A denote the event 'black cube in the first draw', and let B denote the event 'black cube in the second draw'. Pr(A) = ; assuming that each cube is equally likely to be drawn. The outcome of the second draw is dependent on the outcome of the first draw. If event A has happened, there are eight cubes left of which five are black. So: Pr(B I A) = i Now: Pr(A nm = Pr(A). Pr(B I A) 6 5 = 9' Since we are dealing with a two-stage process each with two possible outcomes, a tree diagram could be drawn. First draw Second draw Outcomes Probability 5 B BB ¾x¾= w BW ¾x¾= B WB ¾x¾= 8 w WW ¾x¾= Total Figure 3-16 Note the probabilities in the second draw. They depend on whether a black or a white cube was in the first draw.

27 PERMUTATIONS AND COMBINATIONS 103 Example 21 From a set of five cards numbered 1, 2, 3, 4 and 5, two cards are selected at random without replacement. What is the probability that both are odcvnumbered cards? ' / Number of possible selections = (D = 10 It is favourable if we select two odd numbers and no even numbers when we have three odd numbers and two even numbers to select from. Number of favourable selections = ( ). ( ) = 3. 1 = 3 Required probability = / 0 Alternative method Let A be the event 'odd number first draw' and B be the event 'odd number second draw'. We are asked to find Pr(A nb). The probability that the first card drawn is odd = ¾ i.e. Pr(A) = ¾ The probability that the second card drawn is odd could be either i or¾ depending on whether the first card is odd or even. If the first card is odd, the probability that the second card is odd = 4. 2 i.e. Pr(B I A) = 4 Pr(A nb) = Pr(A). Pr( B I A ) A sample space of the experiment may be considered as consisting of 20 sample points, each equally likely. (1, 2) (2, 1) (3, 1) (4, 1) (5, 1) (1, 3) (2, 3) (3, 2) (4, 2) (5, 2) (1, 4) (2, 4) (3, 4) (4, 3) (5, 3) (1, 5) (2, 5) (3, 5) (4, 5) (5, 4) Note that the ordered pairs (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) are missing. Why? Of these 20 equally likely outcomes, 6 are favourable, namely: So: (1, 3) (1, 5) (3, 1) (3, 5) (5, 1), (5, 3) Pr(A n 6 3 B) = 20 = lo 2 First draw Second draw Outcomes Probability ¾ x¾ =fa Figure 3-17 < o E.- < 3.- E OE 0 EO E EE ¾ x¾ =fa ¾ x¾ =fo ¾ x¾ =fa

28 104 PERMUTATIONS AND COMBINATIONS 5 4 "O "O C: Q) (f) 3 2 Figure First draw 0 Indicates missing points Indicates favourable outcomes 5 Example22 A carton contains 10 transistor, two of which are defective. A sample of three transistors is drawn at random from the carton. Find the probability that not more than one transistor is defective if the sampling is done without replacement. Number of possible selections = ( 1 3 ) = 120 It is favourable if we select no defectives and three non-defectives, or one defective and two non-defectives when we have two defectives and eight non-defectives to select from. Number of favourable selections = ( ). ( ) + (i) G) = = 112 R eqmre. d pro b a b'i' 1 1ty 112 = =15 Alternative method Let D denote 'defective' and N 'non-defective'. First draw Second draw Third draw <D 1 <D 1 9 N < 7 = 1 D 7 10 N 9 6 = = N N D N D N D N Outcomes DON ONO DNN NOD NON NND NNN Probability fo x¼x1 = to ;ix¾xi= Ho = fa x9xa = no = 90 fox¾x½= fto = fa ox.x. = no = 90 foxix¾= m = ox.x. = no = 90 Total Figure 3-19

29 PERMUTATIONS AND COMBINATIONS 105 Note that there is a branch missing. It is not possible to have an outcome DDD. Why? The statement 'not more than one defective' means O defectives or 1 defective. Required probability = Pr(DNN U NDN U NND U NNN) = ill + ill + ill _ Example23 / A group of nine people contains three males and six females. A random sample of five is selected. What is the probability that it contains: a exactly 2 males? b not more than 2 males? a The statement 'exactly two males' implies that the sample of five contains two males and three females. A sample space of the experiment of selecting five pe? ple from nine may be considered as consisting of (;) equally weighted pomts. ( ).( ) points are favourable to the event 'two maies and three females'.. G). ( ) So: Pr(2 males and 3 females) = ---- (;) = ) '3.2.1' Alternative method Pr(first person selected is male) 10 = 21 Pr(second person is male I first is male) Pr(third person is female I first two are male) Pr(fourth person is female I male, male, female) Pr(fifth person is female I male, male, female, female) So: Pr(selecting two males and three females in this order) The problem, however, does not restrict the selection to this order. The two males may be selected in any two of the five selections in ( ) ways. So: Pr(two males and three females) = ( ). J 1 10 = 21