PERMUTATIONS AND COMBINATIONS Factorial Notations The product of all !. For example, Coll the positive integers from I to n is called factorial n and is denoted by the symbol or nl. F 6 1.2. 3.4.5.6 1.2.3.4...20 20 and An Important result nn- 1) (n-2)(n- 3).. 3 .2. 1 =n[ (n-)(n-2) (n-3)...3.2. 1]= n|n- n(n-Dn-2 n(n-)a-2)|n-3 ** Also nn- =n (M-)n-2)..3 .2.1 nn-1 Similarly (n+ 2) (n + DLn =n+2 ote l. Since =nn- n =1 Put -10 1- (1(LO) 0 1 We will also prove at a later stage that 0= 1 Note 2. Factorial of proper fraction and negative integer is not defined. ILLUSTRATIVE EXAMPLES 12 (i) Example 1. Compute o 102 |7 7x6x Sol (0 42 12 (i) 10 2 10x (2x 1) 12 xllx 10 = 66 179 SPECTRUIM DISCRETE t. MATNM AND COMBINATIONS TATIONSAND CONin PHRMTAT 4 n-4 180 .when (i) n= 15, r = 12 4.3.2.-4 2(n-2) (n-39|n-4 (n-2) (n-3) =6 Example 2. Evaluate 6 (n-2) (n-3) (n-2) (n-3) = (3) (2) (i) n-7, (n=6.r=* 6x5x15 Sol () W hen n =6, r=2 n-2 3, or n=5 EXERCISE 1.1 7x6x5x43 7 1Evaluate ()8 8 n 4-. () When n = 7,r =4 34x(3x 2x) 15x14x 13x_12 455 2. Compute 62 (ui) When n = 15, r = 12 15 15 | 12x(3x 2x1) L Compute|4 2. Is|4 [2 [8 find x. when n=5, r=2. Example3. If Evalua |n-r sol when () n=6, r=2 (in n=9, r=5 6. Evaluate -r , find x. 56 64 7 find x. Example 4. If |n+2 =2550 n, find n. 9. If n+1 60|n-1, find n. Sol. +22250 n n+2) (n+ 1) n = 2550 |n +3n-2548 Prove that= =1.3.5.(2n- 1).2". (n +2) (n+ 1)=2550 =0 10. -3t9+10192 -3t10201-3t101 n= 49,-52 12. Basic Counting Principles 2 There are mainly two counting principles namely Rejecting n=-52 as n cannot be negative, we get n =49. () Sum Rule (in Product Rule Example 5. If and are in the ratio 2: 1, find the value of n. 4 n-4 These two principles form the basis of permutations and combinations and so are known as basic cOunting principles. 2:1 Sol. Here 22B- Sum Rule there are two operations such that they can be performed independently in m and n ways, then under of ways in which either of the two operations can be performed is m * n. PECTRUM DISCRETE MA MATIEM AND COMBINATIONSs grample 3. How many 0repetitione sal Given digits are 1, 2, 3, 4, 5 MUTATIONS digit numbers cihe digits is allowed ? G) repetition of the digits is not allowed ? bers can be formed from the digits 1,2,3, 4 and 5 assuming that and when ith can be pertomed ways 183 182 ions Product Rule Product rue is Ako knownm as Fundamental Principle of Counting ie. FPC Fundamental Principle of Counting ie. FPC If one operation can be performed in m different ways and if correspondinp of performing the first operation, there are "n' difterent ways of performine the. number of different ways of performing the two operations taken together is m Extension. If corresponding to each of the m X n ways of performing the two. there are p' diferent ways of performing the third operation then the numi performing the three operations taken together is m Xn Xpand so on. Fundamental Principle of Counting is known as Pundamental Princin Multiplication Principle. If there are wo operations such that one of them can be performed in e. number of given digits = 5 mber of places to be filled 3 are 5 ways of filling ertormod, sawnd operations can be pertormed in n ways, tnen the two operation up the first place, 5 ways for the second place and 5 ways for the each of these the second operation There X n. by fundamental principle of counting, oumber of 3-digit numbers5 X 5 x 5= 125 thid place. ations tak the number of different There are 5 ways wavs of filling up the first place, 4 ways for the second place and 3 ways for the ntal Principle of. third place. Associan total number of 3-digit numbers 5 x4x3 - 60 How many numbers can be formed rom the digits , 2, 3, 9 if repetition of digits is not by fundamental principle of counting, Note We give some example to illustrate the above principle. AMPLES Ermple ILLUSTRATIVE EXAMPLES alowed ? with one digit: There are four digits, hence four numbers of one digit can be formed Example 1. Find the number of 4 letter words, with or witnout meaning, which can letters of the word ROSE, where the repetition of the leters is not allowed. Sol (a) Numbers with the help of these digits. Hence, number of one digit numbers = 4 which can be formed 0u with two digits: First place of two digit number can be filled in 4 ways and the second (6) Numi place can be filled in 3 ways. Hence, number of two digit numbers =4 x 3 = 12. Sol ROSE Number of letters= 4 Number of places to be filled up =4 The first place can be filled up in 4 ways as any one or tne 4 letters can be placed there, Ater up the first place in any one of the 4 ways, there are 3 difterent ways ot filling up the second place one of the remaining 3 letters can be placed there. Iheretore, by the principle of counting, the two taken together can be filled up in 4 x3 ways. After filling the two places in 4 x 3 ways, the third nle be filled up in () Numbers with three digits Number of three digits number = 4 x 3 x2 24. (d) Number with four digits: Number of four digits numbers = 4 x 3 x2 x 1=24. Hence, total number of digits formed with the given digits -4+ 12 +24 +24 64. Example 5. Find the number of different signals that can be generated by arranging at least two flags in order (one below the other) on a vertical staff, if five different flags are available. Sol. Number offlags 5 ways as any one of the remaining two letters can be placed there. So, the thre taken together can be filled up in 4 x 3 x2 ways. After filling the three places in 4 x 3 x2 wa fourth place can be filled up in I ways as the remaining I leter can be placed there required number of words =4 x 3 x 2 x1=24 Example 2. In how many ways can 3 people be seated in a row containing 7 seats ? Sol. First person can be seated in 7 ways A signal can be formed by using two, three, four or five flags. Number of ways of forming signal using two flags =5 x 4=20 Number of ways of forming signal using three flags 5 x 4 x 3 = 60 Number of ways of forming signal using four flags =5 x4 X3x2=120 Number of ways of forming signal using five flags = 5 x 4 x 3 x2 x l= 120 Second person can be seated in 6 ways and the third person can be seated in 5 ways. By the fundamental principle of counting total number of ways in which three persois seven seats in a row S can bex =7x 6 x 5 =210 *1otal numbers of signals formed 20 +60 + 120 + 120320. MATNE SPECTRUM DISCRETE AND COMBINATIONS EXample Find the number of diflerent signals that can be made by amanging at 0na vertical pole, ifo difterent tlags arer available Number of ags o least three flags in class consis ana ureraot hold more than one office? asists of 40 girls and 60 boys. he chosen if the treasurer must be a girl, the secretary In how many ways can a president, vice president, 184 SAND ecretary be chosen , udent may not .ildren are to seated on a bench. 185 n how many wav rangem Sol. Eight chindren are to se the children be seated? and a (a) Number of signas with thre lags In how many ways can the How many aa bench ? ement are possible if the youngest child sits at the left hand end of the First place can e filled in 6 ways Sevond place can he filled in 5 ways ours, how many different signals can be generated if each signai Third place can be filled in 4 ways Gjven 5 tiagS of different use of 2 flags, one below the other? tocal number of signals = 6 x5 x4=120 ways. (6) Number of signals with four hags egure can be made by taking some or all of a number of given things. gement that P: "P, means the number of perm Four places can be filled in 6. 5, 4, 3 way's total number of signals formed = 6 x 5 x4 x3=360 () Number of signab with five flags Meaning of P:P mea Consider three letters a, 6 Pemutoton of permutations of n different things taken r at a time. C. s of three letters taken two at a time are: The permutations of ab, bGca ba, ch, ac he number of arrangements of three Five places can be filled in 6, 5, 4, 3. 2 ways isatoag total number of signalsformed 6 x5 x4x3 x2=720 (dNumber ofsignals with sis Mags Six places can be filled in 6, 5, 4, 3, 2. I ways of three letters taken two at a time is 6 i.e., 'Ph = 6. he total number of signals formed= 6 X 5 x4 x3 Xx2 x I =720 required number of signals = 120 + 360 +720+ 720 1920 Note "Pis also writte itten as P (n, r). A Combination hsa group (or selection) that car tion) that can be made by taking some or all of a number of a given things at a EXERCISE 12 Meaning gor'c,: C,means the number he groups of these 3 letters taken two. at a time are a b, b c, ca. 1. Ram proposes to go to his friend Ali's house n a town which is connected with his will three diferent routes. From there he will go to the city where his uncle lives. The city is con of combinations of n different things taken r at a time. village with the town by two different routes. List out the various possible routes which Ram can o Connece choo 2. If there are 20 steamers playing between places A and B, in how many ways could the round Mhstration. Consider three letters a, b, c. from his village to go to the city. 1s group e concerned a c or ca Is the same group, as in a group we are concerned with the from A be made if the returm was made on s far 85 fained, whereas in the case of arrangement we have to take into consideration the () the same steamer, (i) a different steamer ? r of things contai arder of things Nate. 'C, is also written as C (n, r). 3. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there How many 4-leter codes can be formed using the first 10 letters of the English alphabet, if mf:We know that the number or permutaions ot n different things taken r at a time is the same as the letter can be repeated? 5. How many 3-letter code words are possible using the first 10 letters of English alphabet if nd the number of permutation fn different things taken r at a time ie., find the value of "P. (0 no letter can be repeated? (i) letters are repeated? rof ways in which r places in a line can be filled up with n persons. Re fist place can be filled up in n diferent ways as any one of the n persons can be placed there, tr ling up the first place in any one of n ways, there are (n - 1) different ways of filling up the second 6. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each numbe,y one 0 unc Tenang - 1) persons can be placed there. Therefore, by the princinle of BC ition,the two places taken together can be filled up in n (n-1) ways. After the two places have been id u n any of the n (n - 1) ways, the third place can be filled up in (n -2) ways as any one of the How many 3-digi even numbers can be formed from the digis 1, 2, 3, 4, 5, 6 if the dugis can h-2) persons can be placed there. Therefore the three places taken together can be filled up in starts with 67 and no digit appears more than once? repeated? nb -1)1-2) ways. The factors hegin with n and go on diminishing by unity. th tactorn- D -r*l redinE in this was wr ser that henever a placr is filled up a new factor is introduced. ..ATIONS AND BINATIONS AP, - 6. P, 2. Find r, if s Tample 5. P,6. P, +1) 187 n(n-Dn-2)... (n-r4 (n- (n-r)..3.2.1 15--1 rumbr et navs of filling up r places n("- D7-2)... (n-p+i 6 (-1)(n - 2)... (n - r+ 1) = (6-r) (5-)|4- 4- 6 (6-) (5- ) (6- r) (5 - r) = 6 - 11r+24 =0 r- 8) (r-3) =0 Cor. We know that "P, = n (n - 1) (n-2)... (n-r+ 1) r8, 3 ecting r = 8 we get, r=3 Find n if P(9, ,5) + 5P(9, 4) = P(10, n) and mple 3. Ea P (9, 5) + 5 P(9,4) = P (10, n) r=n in (1) and (2). Pn(n- )(n-2)... (n-n+ ) n(n- 1)n-2). I = Putting Ps+5 °P= P, sol. 10 10x9 10-n 9-5 P _10 10-n and 4 0 10 10-n I: 4-24) From (3) and (4). we ge. 0 12 10-n 10-n= 120 10-- 10-n= 5 5. ILLUSTRATTVE EXAMPLES Examp 4. Find rif P(l0, r+ 1):P(11,r)= 30: 11. Sol. Since P(10, r + 1): P(11, r) =30 : 11 Example 1. Prove that OP+1"P, = 30: 111 "P, = 2. "P,-2 ( P = P +3 x P Tp, 10 |10- r-1 Sol. () LH.S. = "P, B0 30 11 R.H.S. 2 "Pn-2 ^ [n-(n-2) 11 L.H.S. = R.H.S. (in) L.H.S.= "P, = 10 x 9 x8=720 R.H.S.="P+3. "P=9 x 8 x7+3 (9 x 8) = 504+216 720 30 19- llx10 10 (11-)(10-r)9-30 9- (11-r) (10-r)=6 X5 LH.S. = R.H.S. 11-r=6 r=5 SPECTRUM DisCRE COMBINATIONS 1ONS AR aCe each word is to end withi ( Smce cach wor fix E in the end th E EXERCISE 1.3 189 are to be arranged in 7 places ind n if Rn, 4) 0Pn, ) und n it P n ) 100 P (n. ) 7 letters are lo be arranged rof words= 'P;= 2 7 x6 X 5 x 4 x 3 x 2 1 = 5040 required number of words = ch word is to begin with T and end with E tind n it 0 Pn, 6) = P(n 2. 7). (ii) Since cach word n>4 fix Tin the beginning and E in the end t urkd the value of n sUch that P arranged in 6 places 6 letters are to be a ()P 4 "P;. n>d required number of words = P,= many words (with or without dictionary meaning) can be made from the letters in the P 6 =6 x 5 X4 x 3 x2x 1 = 720 5. Find r if: Erample Ho ONDAY, as5suming t ()P, = "P- no letter is repeated, if ( P =2 °P- MON (ii) all letters used at a time 6 Find n if 'P;: P, = 1:9. 4 letters are used at a time all letters are used bu AY. but the first is a vowel ? Ir P- P,+3 = 30800: 1, find r. m) 1.6. Practical Problems involving Perm utation Now we will apply the formula sol. The given word is MONDA Y. Number of letters = 6 umber 6 of letters to be taken at a time = 4 to practical problems. 9 Numt P(n r) =n (n - 1)(7-2)... (7-* i) of words = "P4 =6 x5 x4 x 3 = 360 required number of words = (i) Number of letters to be taken at a time = 6 require number of words =P 6 x 5 x 4 x3 x 2 x 1 = 720 ILLUSTRATIVE EXAMPLES Example 1. How many 3-letter words can be made using the letters of the words ORIT ORIENTAL? Sol. Given word is ORIENTAL 2 vowels 0 and A can be filled in 2 ways. ii) First place with 2 number of letters =8 Number of letters to be taken at a time =3 Now remaining 5 places can be filled up with 5 letters in P ways. required number of words = °P3 required number of words =2 x P 8x7x6=336 = 2 x (5 x 4 x 3 x 2 x 1)=2 x 120 240 nle 4, It is required to seat 5 men and 4 women in a row so that the women occupy the even places. Example 2. Find the number of different 8-letter words formed irom tne letters of the word TRIANG Example 4. It Is required to seat 5 men and 4 women in each word is to RIANGU () begin with T (i) end with E (ii) begin with T and end with E. Sol. The given word is TRIANGLE How many such arrangements are possible ? Sol. Number of men = 5 Number of letters = 8 Number of women = 4 Number of letter to be taken at a time = 8 Nine places are to be filled. In these 9 places, only four are even. Therefore, 4 even places can be led with 4 women in 'P, = 4 ways. Aiso remaining 5 places can be filled up with 5 men in () Since each word is to begin with T fix T in the beginning letters are to be arranged in 7 places P,= Sways. required numbers of words = P,=|7 7 x 6 x 5 x 4 x 3 x2 x1 = 5040 required number of arrangements |4 x[5 (4 x 3 x2 x 1) x (5 x 4 x3 x2 x ) 24 x 120 2880 sPECTRUM DIscr RETE MA Evampe 5. Find the number of different 8-lctter words formed from the lettere Cach word is to have voncls orupy ing Odd paces. PERETATIONS How many numbers lying nle of the digits is not allowed? of t AND COMBINATIONS rs word TRIANG 190 betwe 100 and 100 Erampl rthe repetition of the E between T00 and 1 000 consis of three digits. 191 can be formed with the digits 0, 1,2.3. 4,5. Sol. R Given digits are 0, umber of given digits =6 2 sol. Numbers between s are 0, 1, 2, 3, 4, 5 marked places in "P3 ways. .mber of digits to be taken at a time = 3 owels ocupy odd places numbers formed of three digits= P3 =6 x 5 x 4 = 120 the three vowels can be arranged in four Also the five consonants can be arranged among themselves in S wayys eis But these include those and not of three dig numbers which have 0 on their extreme left and these will be numbers of 2 4x 3x 2 x1) =24 x 120 = 2880 = (4 x 3 x2) x (5 x 4X 3x 2 x1) = 24 x 120 aning, can be formed using all the let required number of words = "P; X[5 digits. Example 6. How many words, with or without meaning, can be formed using EQUATION at a time so that the vowels and consonants occur together ? Sol letters of the w we have to exclude these from the total. which have 0 on the extreme left position, we fix 0 in that position and fill up nbers In order to find numbe aut of five digits at our disposal which can be done in °P, ways ie. 5 x 4 ie EQUATION Consonants are Q. T, N the remaining two places Vowels are E. U, A. I, O 20 ways. required numbers=120- 20 = 100 so How many ddnumbers greater than 80000 can be formed using the digits 2. 3, 4, 5 and 8 if Consider 3 consonants as one letter and also 5 vowels as one letteer. Erample. it is used only once in a number 2 two letters can be aranged in 2 ways. Also 3 consonants can be arranged among themselves in |3 ways and 5 vowels in s inSol. Given digits are 2, 3, 4, 5, and: number of given digits = 5 Number of digits to be taken at a time = 5 mselves in and ways. required number of words 2 X *B (2x 1)x (3 x 2 x 1) x (5 x 4 X 3 * 2 x 1)=2 x 6x 120- Since number is to greater than 80000 Example 7. In how many ways can 5 girls and 3 boys be seated in a row so that no two bo: Sol. Let 5 girls be G, G2. G3, G4. Gs ogether. first digit from left should be 8 x G, x G, x G3 x G4 x Gs X fix 8 in the beginning. no two boys are together Now 4 places are to be filled with 4 digits. 3 boys can be arranged in 6 "X marked places in P3 ways. Again numbers are odd numbers should have either 3 or 5 in the end Also 5 girls can be aranged among themselves in5 ways. required number of ways= °P3 X5 end's place can be filled in 2 ways. Example 8. How many 4-digit numbers can be formed by using the digits I to9 if repetition of digiti not allowed? (6 x 5 x 4) x (5 x 4 x 3 x 2 x 1)= 120 x 120 14400 Remaining 3 places with 3 digits can be filled in 3P3 ways required numbers =1 x2 x *P3 = 1 x 2 x (3 x 2 x 1) = 12 Sol. Digits are 1,2. 3, 4, 5, 6,7, 8,9 Example 11. How many different signals can be formed with five given flags of different colours ? total number of digits =9 Number of digits to be taken =4 Sol. Number of flags= 5 numbers formed= P4 =9 x 8 x 7 x 6 3024 A signal may formed by hoisting any number of flags at a time. Number of signals by hoisting one flags at a time = °Pi Number of signals by hoisting two flags at a time = P2 sPECTRUMDISCRETE M ONS TIWEMN ONS AND COMBINATION af different 8-letter words formed from the letters of the word TRIANGLE if each word IS to have vo ays can 4 b w many ways ind the nunmber have vowels c occupying the second, third and fourth places. boys and 3 girls be seated in a row of 7 chairs if boys and girls alternate? 193 Nunber of sznals by hoisting four tlags at a time= P Numer of sgnals by hoisting five flags at a time = P total numher of sgnals tomed PPP: P+ P S5N4-sx4X3+5x4X3 X2+5 x4 x3x2: 192 Number of signals by hoisting three tlags at a time= P, periods in each at each subject is allowed at least one period ? the letters of the wordPENCIL' be arranged so that N is always next to E? ch working day of a school, in how many ways can one arrange subjects such that ea can f there are 3 in how many way d the 2 x1 rent 8-letter words formed from the letters of the word TRIANGLE if umber of differen have consonants never together. of different 8-letter arrangements that can be made from the letters of the word Find. each word is to have cons I Find. JGHTER so that all 20 0- 120 120 = 325. the number vowels never occur together. EXERCISE 1.4 come in the first ys can 5 bo ame subject remain together? books on Chemistry and 4 books on Physics be arranged on a shelf so 16 In how many way 1. Ten horses are nunning a race. In how many' ways can these horses. that the books on same (i) Seven songs are to be rendered in a programme. in how many differene ain office in a municipal ele 8 students appearing remaining 5 in differ in an examination, of which 3 have to appear in a mathematics fferent subjects. In how many ways can they be made to sit in a could they ramme. In how many different orders. I7There paper and the an third place, assuming no ties in mathematics cannot sit next to each other? (ui) There are sin candidates contesting for a certain oftice in a municinal ways can their names be listed on a ballot ? Tendered ? rOw. if the candidates 18 Inho In how many ways can 4 b I9 Three hree married couples ar he seated next to each other, in how man can 4 boys and 3 girls be seated in a row so that two girls are together? nles are to be seated in a row having six seats in a cinema hall. If spouses are to each other, in how many ways can they be seated ? Find also the number of ways oW t (n) Four books, one each in Chemistry, Pnysics, Biology and Mathematic. a shelf. In how many ways can this be done ? are to be artange burs if eachs Find the number of c of their seating if all the ladies sit together. (1) How many different signals can be generated from 6 flags of different makes use of all the flags at a time, placed one below the other ? umber of different 8 letter words formed from the letters of the word TRIANGLE if 2. How many words, with or without meaning. can be rormed using all the let EQUATION, using each letter exactly once? he formed using all he letters of the () have no two vowel together (Gin have both consonants and vowels together each word is to 3. There are 6 items in column A and 6 items in column B. A student is asked to mat column A with an item in column B. How many possible answers (correct or incorect match each t 1. How many 3-digit numb Uew many different 4-digit numbers can be formed from the digits 2, 3, 4 and 6 if each digit is then (i) have the relative position of the vowels and consonants unaltered. nbers can be formed by using the digits I to 9 ifno digit is repeated ? the question? 4. (a) From a committee of 8 persons, in how many ways can we choose a chairman chairman assuming one person cannot hold more than one position ? uSed only once in a number ? Further, how many of these numbers (a end in a 4? (i) end in a 3 ? (ii) end in a 3 or 6? and vi many ways can we select president, vice presite 13, Find the number or4-01git numbers that can be formed using the digits secretary anda treasurer if each of the 12 candidates can hold any office ? Ha manv diferent 5-letters words can be formed out of the letters of the word "DELHT ?54. How many difterent numbers between 100 and 1000 can he fom ,2,3,4, 5 if no digit is repeated. How many of these will be even? many of these will begin with D and end with E ? and 6, assuming that in a number, the digits cannot be repeated ? How many of these will be divisible by 5? 3 How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated? % How many 4-digit numbers are there with no digit repeated ? 6. The leters of the word TUESDAY are arranged in a line, each arrangement ending with lete! How many different arrangements are possible ? How many of them start with letter D ? 7. Find the number of different 8-letters words formed from the letters of the word TRIANGL each word to have T and E at the end places. 8. Find the number of different 8- letters words formed from the letters of the word Eyun gvEn numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? each word is to start with a vowel. 21 9. In how many ways can 6 boys and 5 girls be arranged for a group photograph if tne S on chairs in a row and the boys are to stand, in row, behind them ? 4 How many numbers greater than 40000 can be formed using the digits 1, 2, 3, 4 and 5 if each digit s are n 5 Used only once in each number ? Row many of the natural numbers from I to 1000 have none of their digits repeated? WTATIONSAND COMBINA how many ways SPECTRUM DIsCRET SAND COMBINATIONS them are Ake and s can the letters of the word 195 ASSINATION be aranged so that all the 9 'a,q ofth d ef.. hem being ASSASSINATION letters = 13 17. Find the number of permutations of n things taken all at a time whenp afss kind, g of them are alike and ofa second kind, all others being difèrent. mutations and act p Number of A 's =3 =4 ber of given leters'a' by p' diferent letters aj, 4p. Ihese new p leters can be arrano new arangements, when a's are considered diferent. And, there fore, if such permutations, then the total number of pemutafions will be r X |p. Proof. Let n things be denoted by n letters, p of them being alike and denoted by arranged among themseva Number of S's= Nuntber of l's =2 and denoted by 6' and the remaining being all diferent and denoted by c, d if such a change is n made in al Letr be the required number ofpermutations. Take any one of these permutas. Number of N 's =2 r the four S's as c replace q like letters by ferent one letter b,.6 These g letters can be arranged among themselves in g ways. one such permutation will give rise to lg pemutation and if such a onsider Now, consider one of these r Xp amangements, and replace 10 Ich a change is made to allte 10 letters can be arranged in 22 , arranged among themselves in ways. 2 pemutations then the total number of permutations will be r X[p X|q But number of permutations of n diferent things taken all at a time is = |n requir red number of ways in which four S's are always together ' e have xpx\e=e 10 third Ertension. This rule can be extended if in addition to the above r things are alike an 10x9x8x7x6x5x4x| 3 |4 3x(2x l)x(2x L3212 x1 =151200 and so on. Emple Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of ILLUSTRATIVE EXAMPLES these arangements 0 do the words start with P c do all the vowels always occur together Etample 1. Find the number of permutations of the letters of the word ALLAHABAD. Sol ALLAHABAD (i) do the vowels never occur together (o) do the words begin with I and end in P? Total number of given letters =9 Number of A's =4 l NDEPENDENCE Number of L's =2 Number of given letters = 12 9x8x7 1x6x5 - 7560 required number of words= Number of N's =3 42 4x(1x 2) Number of E's = 4 Example 2. In how many ways can 5 flags, in which 3 are red, one is white and one is blue, be arrangei a staff, one blow the other, if flags of one colour are not distinguishable ? Number of D's = 2 Sol. Total number of flags =5 Number of red flags =3 Number of white flags=1 Number of blue flags 1 required number of arrangements = |12 42 12 x11x10x9x8x7x6XD*= 1663200 required number of arrangements= 5x4x3 3x2xl)x[4x (2x 1) Since each word is to start with P 20 fix P in the beginning. RETE MAT sPCTRM DISCRET TIONS (ONS ANDCOMBINA TIO /11 EXERCISE 1.5 RTATONS A ND 197 How manY permutations. In how many ways can 4 re. rmutations of the letter of word APPLE are there ? avs can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the o 138600 Remaung Il letters can de arranged in, same colour are indistinguishable 2 There are. white, 4 w. Assuming that all the s of same colou red and blue marbles in bag. They are drawn one by one and arranged in a 8 marbles are drawn, determine the number of different arrangements if (2D/4N(2A ) lour are indistinguishable. (7) Consder tive voue/s as one Ictter many distinct ways can the product xy be written without using exponents lifferent ways, the letters of the word ALGEBRA can be arranged in a row if marb In how ermutations of the letters in MISSISSIPPI do the four I's not come together? leners can e arranged in Was ways. In how many distis 6. In how two A 's are together ? in the two A's are not together ? number of arrangements in which vowels are always together 8x 7x6x5xdx|3 Aso five vowels can be arranged in themselves in the tw Find how many arrangeme how many of them n consonants occur together aw many arrangements can be made with the made with the letters of the word 'MATHEMATICS ? In 5X 3360 X 5= 16 occur together (i) vowels do not occur together ? emiutations of all the letters of the word EXAMINATION are listed as in a nany words are there in this list before the first word starting with E 2 3x (lx2) If the diffe dictionary, how many Words are there i how many ways can the (it) Required number of arrangements = 1663200 -16800 = 1646400 () Fix I in the beginning and P in the end. 10 In hou words start with P and end with S, (i)vowels are all together letters of the word PERMUTATIONS be arranged if the required number of arrangements E42 vays 4 letters between P and S ? 10x9x8x 7x 6x5x| 4 (3x2x1)x4x(2x 1) (i) there are alwa How many 5-digit « 12600 even numbers can be formed using the digits 1,2, 5, 5, 4? = 10. 1.8. Cireular Permutations d the number of ways in wh. Example 5. How many numbers greater than 1000000 can be tormed by using the dioite which n persons can be arranged at a round table. orsons are sitting around a circular table, then there is no first and last person. Let us fix Find e ,242 SoL. Given digits are 1, 2,0,2,4,2,4 Proof. ne nerson. The remaining (n - ) persons can now be arranged in the remaining (n 1) total number of digits = 7 the position of one person. The. '. places in Pn-1 Le, n ways. required number of ways = n-1. 19. Clockwise and Anti-clockwise Permutations The total number of circular permutations can be divided into two types Number of 2's =3 Number of 4's =2 Number of digits to be taken at a time = 7 7 numbers formed 9 Clockwise In two such arrangements each person has the same neighbour though in the reverse order and either of these arrangements can be obtained from the other by just over-turning the circle. If in this case, no distinction is made between clockwise and anti-clockwise arrangements then the two such arrangements are Considered as only one distinct arrangement. 420 3x(1x2) (i) Anti-clockwise. These numbers also include those numbers which have 0 at the extreme left position. Vumbers having 0 at the extreme left position 6x5x 4x 3 2 3(142) required number of numbers = 420 60 = 360 Hence the number of circular permutations in such cases Note. Questions on necklaces with beads if different colours are to be tackled by the above formula, sn this case also there is no difference between clockwise and antiwise arrangements. SPECTRUM DISCRET E MATWEN NS AND COMBINA TIONS 199 mber of combinations of n dissimilar things taken r at time (by using the value of "P,). 198 PERMUTATIONS, Find the required number of combinations of n things taken r at a time be x. LLUSTRATIVE XAMPLE : Let the required each mbination contains r things, which can be arranged among themselves in ways. coresponding onding to each combination, we get permutations. colours fom a necklace Example l. In how many ways can 6 beads of different c Sol. Number of beads = o due tor combinations = x total numbero permutations s ofn things taken r at a time = "Pp. Trequired number of nevklaces In how many ways can de But number of permutations Exampe 2 Four persons A. B. C and D are to be seated at a circular table or we have, seated ? Sol Number of given persons = 4 ular table =|4-1 =33 x2 x1 =6 Number of ways of seating them in a circular Example 3. In how many ways can 5 boys and 5 girls be seated at a round table together SoL Let the boys be seated firstiy leaving one seat vacant in between each of two boVs s can 5 boys and 5 girls be seated at a round table, so that no t n(n-1 (n-2).. (n-+) [: no two girls are to sit toget L12.Complimentary Combination Prove that C= "Ca-r This can be done in 5-1 i.e., in |4 ways. Now S girls can be arranged in 5 vacant seat Ps ie., 5 ways. reguired number of ways = |4x5 LHS- Proof. = (4 x 3 x 2 x 1) x (5 x 4 x 3 x2 x 1) =24 x 120 =2880. EXERCISE 1.6 R.H.S.=C-rn- |n-(n-) in how many ways can 8 girls be seated at a round table provided Parveen and Vipul are not n together? L.H.S. = R.H.S. Another method. (Fromfirst Principles) in how many ways 4 boys and 4 girls be seated at a round table provided each boy s between two girls? We know that C, means the number of combinations ofn things taken r at a time. Whenever we form grup ofr things out of n things, (n - r) things are left which themselves form a group. 3. The Principals of six colleges seat themselves round a table to discuss the studen proDIem. in how many arrangements, Principals of X-college and Y-college will nor sie ken (n-7) at a time. number of groups of n things taken r at a time is the same as the number of groups of n things 1.10. Combination ie, "c-"Cp In combination, we are not interested in arranging, but only in selecting r objecis no fact, we do not want to specify the ordering of these selected objects. oe. We will aply this formula in problems in whichr> Differences between a permutation and a combination L3. Prove that c,+"C="*'C is mau rol. [From first Principles O Only a selection is made whereas in a permutation not only a selection id also an arangement in a definite order. 2. In a combination, the ordering of the selected objects is immaterial whereas in a permulatio We know that, atio, u ordering is essential. n*C= humber of combinations of (n + 1) things taken rat a time sAND COMBINATIONS SPECTRU DISCRETE MATHEMAT LLUSTRATIVE EXAMPLES 200 number of combinations which include a particular thing that C (8, 4) + C (8, 3) = C (9, 4). number of combinations which exclude that particular thing. L.H.S=C(8, 4) +C (8, 3) ="C, + C 8x 7x6,8*7x6 704 Ix2 x 3 x 4 1x 2 x 3/0+ 56 = 126 fumple number of combinations of n things taken (r- I) at a time. number ot combinations of n things taken r at a time. c, - "C+ C R.H.S.- C(9,4)= "Cs1x2x X8x7x 3x4 6 126 C. c."C Another meth LH.S. =R.H.S. n Eample. Evaluate sc CC sC, C (15, 14). R.H.S Here "C,="C,--] 50 x 49 x 48 Ix 2 x3 LHS "C,+C 19600 = |r-1.(n - r+). - rl1|n- Again C (15, 14) = "C="C, I:'c="C- r+n-r+ r-1n-rlr(n-r+ 15 Eumple 3. Determine n if (n+1 rT7 (n-r+1 -D(n-r+ Dn=r kh-r n+1 2C: "C3 11:1 (2 n) (2n-1) (2n-2) LH.S. R.H.S. 1.2.3 1.14. Find the total number of combinations of n different things by taking some or all at a time. OR n (n- 1)(n -2) 1.2.3 2n) (2n-1) (2n-2) 11 4(2n-)(n-)_1 n(n-1) (n- 2) Find the value of "C+"C,+"C+.+ "Cn (n- 1) (n -2) 1 Proof: Here each thing can be dealt in two ways. Either it is included or excluded from a selecto 4(2n-1) 1 Hence each thing can be disposed in 2 ways. n-2 11n-22-8n-4 Dur each way of dealing with a thing is associated with each way of dealing with the otner total number of ways of dealing with n things 3n= 18 n6 ample 4.If "C = "Cg, find "C17 2 X 2x 2 x .....n items = 2 Here "C "CR But these include the case where none of the things is selected. either 9 8 the required number of ways = 2"- 1 which is impossible or n=9+8 ga, we may select I or 2 or 3 or 4...or n things out of the given n dififerent tnin n= 17 C-C1 the required number of ways ="C+"'C,+ "C3 *...+ "C, =2"-1 SPECTRUM DISCRETE MAT HEMATO TIONS AND CoMBINATIONS RWAT1ONS AND 202 x 203 mple 5. Find the vaBur of "c, C - -r- n- n-r- ol Consider C. C C *c *C; *C,+ "C -°'CG+"G)+ "c + "C+ "C :"C-1 + "C, = "*in 3n-3r= 2n n 3r - ("c. - " c:)- "C, +°C, + S" C C S2S1x 50 x - 270725 and ntlC n-r+1 Eampk 6 If C,= S6 and "P 336, find n and x. 56 (-r+1)| n- (n+1) Sol C 56 Lzn-z 336 n-r+1 P- 336 Dividing (2)) by (1), we get, n-r+l n+1 x 336 13n-13 r+ 13 =9n+9 6 4m-13r+4 =0 4(3)-13 r+4 =0 : of(1)] "P 336 12r-13r+4 =0 Now "P, = 336 -r+4=0 r=4 n (n- 1)(n-2) = 8x 7 x6 n=8 from (1), n=3 x 4= 12 we haver =3, n =8 n= 12, r=4 Example 7. If c, : "C, : "*'C, =6:9: 13, find n and r. EXERCISE 1.7 Sol. We have n-CC, : "*'c, -6:9: 13 1. Evaluate C(19, 17) +C (19, 18). Prove that 0 1+C(3, 1) +C (4. 2) = C(5,3) ) C(2,1) +C(3, 1) + c (4, 1) = C 3, 2) +C (4, 2) Delermine n if 2nC: "C2 = 12: 1. n-r 4If "C= "C2, find "C2 - .MAT oYS AND COMBINATIONS SPECTRUM DISCRETE 205 af combinations of 5 cards out of 52 : = "C= 52 x S1 50 49 48 2 3x 45 cards containing no king = ""C, = - C 10) C, 1), determine n and hence C (n, 5). 48x 47x 46 45 44 Number of combination I23 45 C C-t. tind r. equired number of combinations PrOve that C7, n) 52 x 51x 50 x 49 x 48 48x 47x 46 x 45 x 44 Ix 2x 3 x4x5 Ix 2x 3x 4 x5 = 886656 tm= C(n. 2) prove that C(m. 2) =3 C (n + 1, 4). ays can we select a cricket eleven from 17 players in which only 5 players can If C: C-, =1:2 and C, "C =2:3, find n and r. nnle 4. In how many ways can we sele include exactly 4 bowlers ? ow/if each cricket eleven must o Number of bowlers =5 Number of other players = 12 7 other players out of rs Qut of 12 other players can be selected on "C7 ways and 4 bowlers out of 5 bowlers .15. Practical Problems lnvobving Combinations ILLUSTRATIVE EXAMPLEs Eaaple 1. In how many ways cana committee be selected from 15 persons if the (i) 13 members. selected from 15 persons iT the committee is to h ways. () members. can be selected in C required number of teams= *C7 x °C4 = C, x 5c,= 2xllx 10x9x8 6 1x2x3x4x5 1 e5, A commitee of 3 persons is to be constituted from a group of 2 men and aVS can this be done ? HoW many or these committees would consist of I man and 2 women ? SoL Total number of persons = 15 ()Number of persons to be selected =3 15x14xl = 455 C1x2x3 3960 required number of ways= C women. In how () Number of persons to be selected = 13 Number of men =2 Sol () 15x14 x2 Number of women =3 required number of ways = C; ="C2 = =105 Example 2. In how many ways can a student choose a programme of 5 courses if 9 courses are availst and 2 courses are compulsory for every student total number of persons =2+3 5 Number of persons to be taken in a group =3 = 9 C required number of committees =C3 = °C2 Sol. Total number of courses Number of courses to be chosen = S5 Sx4 = 10 two courses are compulsory Ix2 student is to select 3 courses out of 7 courses ) I man out of 2 men can be selected in C, ways. 2 women out of 3 women can be selected in required number of ways = "C, = 35 Ways. Example 3. Determine the number of 5-card combinations out of a deck of 52 cards if alleast five cards has to be a king required number of committees Men Women -2C C2 Sol. Total number of cards 52 3 Number of king remaining cards 4 48 2 gMITA TIONSAND COMR 4 red cards out of 26 SPECTRUM DISCRETE M ATWEMAT AND COMBINATIONS Esmpk har s the number of ways of choosing 4 cards from a pack of 52 , of these playi cards ? lh 20 of 26 red cards can be selected in °C4 ways. Also 4 black cards out of 26 selected in C ways hiat cards can be reguired number of ways = 4°C, + 26 bCs tor cards are of the same suit 2 x c2 x 26x25x 24x 23 -29900 tu fourcards belongs to four differet suits, Ix 2x3x4 consists ncists of 4 girls and 7 boys. In how many ways can a team of 5 members be (a)are facecards ample 7. A group geted if the team has (nogirls ? (i) at least three girls ? (7) wo are red cards and rwo are black cards, (v)cands are of the same colour? (in at least one boy and one girl ? S Total mumber of given cards 52 Number of cards to be selected at a time =4 52x51x 50x 4 -270725 s Number of boys = 7 Number of girls =4 Ix2x 3x4 Boys Girls required number of ways=*C 4 cards in eachs suit Committee is to consist of 5 There are four suits namely diamond, club, spade and heart. There are 13 CWay0 committee includes no girl we are to select 5 boys out of 7 boys ds, Cways of choosing 4 clubs, 13c There areC, ways of choosing 4 diamonds, choosing 4 spades and "C ways of choosing 4 hearts. 7x6 number of committees required number of ways = "C + C +"C, + 13c, 13x12xl1x10 - 2860 -4 x 3C4 X 12x3x4 (a committee is to include at least one boy and one girl different possibilities are There are four suits and 13 cards in each suit. () (o 1 boy, 4 girls Boys Girls (6) 2 boys, 3 girls There are C ways of choosing I card from 13 cards of each suit. () 3 boys, 2 girls by the multiplication principle, 4 4 boys, 1 girls required number of ways = "Cx cxC, x "C, = 13 x 13 x 13 x 13 = 28561 required number of ways (i) Jacks, queens and kings are face cards -'Cx c +"C2 x *c3 *+'Cg x °C2 + 'C4 x *C total number of face cards = 12 -'C x1+ 'c3 x *c+ 'c, x *c2 + 'C x *c Number of cards to be selected at a time = 4 7x6x5 4x3 1X1 x I2x 7x6xS required number of ways = 12c