Chapter 7 PERMUTATIONS AND COMBINATIONS Every body of discovery is mathematical i form because there is o other guidace we ca have DARWIN 7.1 Itroductio Suppose you have a suitcase with a umber lock. The umber lock has 4 wheels each labelled with 10 digits from 0 to 9. The lock ca be opeed if 4 specific digits are arraged i a particular sequece with o repetitio. Some how, you have forgotte this specific sequece of digits. You remember oly the first digit which is 7. I order to ope the lock, how may sequeces of 3-digits you may have to check with? To aswer this questio, you may, immediately, start listig all possible arragemets of 9 remaiig digits take 3 at a time. But, this method will be tedious, because the umber of possible sequeces may be large. Here, i this Chapter, Jacob Beroulli (1654-1705) we shall lear some basic coutig techiques which will eable us to aswer this questio without actually listig 3-digit arragemets. I fact, these techiques will be useful i determiig the umber of differet ways of arragig ad selectig objects without actually listig them. As a first step, we shall examie a priciple which is most fudametal to the learig of these techiques. 7.2 Fudametal Priciple of Coutig Let us cosider the followig problem. Moha has 3 pats ad 2 shirts. How may differet pairs of a pat ad a shirt, ca he dress up with? There are 3 ways i which a pat ca be chose, because there are 3 pats available. Similarly, a shirt ca be chose i 2 ways. For every choice of a pat, there are 2 choices of a shirt. Therefore, there are 3 2 6 pairs of a pat ad a shirt.
PERMUTATIONS AND COMBINATIONS 135 Let us ame the three pats as P 1, P 2, P 3 ad the two shirts as S 1, S 2. The, these six possibilities ca be illustrated i the Fig. 7.1. Let us cosider aother problem of the same type. Sabam has 2 school bags, 3 tiffi boxes ad 2 water bottles. I how may ways ca she carry these items (choosig oe each). A school bag ca be chose i 2 differet ways. After a school bag is chose, a tiffi box ca be chose i 3 differet ways. Hece, there are 2 3 6 pairs of school bag ad a tiffi box. For each of these pairs a water bottle ca be chose i 2 differet ways. Fig 7.1 Hece, there are 6 2 12 differet ways i which, Sabam ca carry these items to school. If we ame the 2 school bags as B 1, B 2, the three tiffi boxes as T 1, T 2, T 3 ad the two water bottles as W 1, W 2, these possibilities ca be illustrated i the Fig. 7.2. Fig 7.2
136 MATHEMATICS I fact, the problems of the above types are solved by applyig the followig priciple kow as the fudametal priciple of coutig, or, simply, the multiplicatio priciple, which states that If a evet ca occur i m differet ways, followig which aother evet ca occur i differet ways, the the total umber of occurrece of the evets i the give order is m. The above priciple ca be geeralised for ay fiite umber of evets. For example, for 3 evets, the priciple is as follows: If a evet ca occur i m differet ways, followig which aother evet ca occur i differet ways, followig which a third evet ca occur i p differet ways, the the total umber of occurrece to the evets i the give order is m p. I the first problem, the required umber of ways of wearig a pat ad a shirt was the umber of differet ways of the occurece of the followig evets i successio: (i) (ii) the evet of choosig a pat the evet of choosig a shirt. I the secod problem, the required umber of ways was the umber of differet ways of the occurece of the followig evets i successio: (i) (ii) (iii) the evet of choosig a school bag the evet of choosig a tiffi box the evet of choosig a water bottle. Here, i both the cases, the evets i each problem could occur i various possible orders. But, we have to choose ay oe of the possible orders ad cout the umber of differet ways of the occurece of the evets i this chose order. Example 1 Fid the umber of 4 letter words, with or without meaig, which ca be formed out of the letters of the word ROSE, where the repetitio of the letters is ot allowed. Solutio There are as may words as there are ways of fillig i 4 vacat places by the 4 letters, keepig i mid that the repetitio is ot allowed. The first place ca be filled i 4 differet ways by ayoe of the 4 letters R,O,S,E. Followig which, the secod place ca be filled i by ayoe of the remaiig 3 letters i 3 differet ways, followig which the third place ca be filled i 2 differet ways; followig which, the fourth place ca be filled i 1 way. Thus, the umber of ways i which the 4 places ca be filled, by the multiplicatio priciple, is 4 3 2 1 24. Hece, the required umber of words is 24.
PERMUTATIONS AND COMBINATIONS 137 Note If the repetitio of the letters was allowed, how may words ca be formed? Oe ca easily uderstad that each of the 4 vacat places ca be filled i successio i 4 differet ways. Hece, the required umber of words 4 4 4 4 256. Example 2 Give 4 flags of differet colours, how may differet sigals ca be geerated, if a sigal requires the use of 2 flags oe below the other? Solutio There will be as may sigals as there are ways of fillig i 2 vacat places i successio by the 4 flags of differet colours. The upper vacat place ca be filled i 4 differet ways by ayoe of the 4 flags; followig which, the lower vacat place ca be filled i 3 differet ways by ayoe of the remaiig 3 differet flags. Hece, by the multiplicatio priciple, the required umber of sigals 4 3 12. Example 3 How may 2 digit eve umbers ca be formed from the digits 1, 2, 3, 4, 5 if the digits ca be repeated? Solutio There will be as may ways as there are ways of fillig 2 vacat places i successio by the five give digits. Here, i this case, we start fillig i uit s place, because the optios for this place are 2 ad 4 oly ad this ca be doe i 2 ways; followig which the te s place ca be filled by ay of the 5 digits i 5 differet ways as the digits ca be repeated. Therefore, by the multiplicatio priciple, the required umber of two digits eve umbers is 2 5, i.e., 10. Example 4 Fid the umber of differet sigals that ca be geerated by arragig at least 2 flags i order (oe below the other) o a vertical staff, if five differet flags are available. Solutio A sigal ca cosist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us cout the possible umber of sigals cosistig of 2 flags, 3 flags, 4 flags ad 5 flags separately ad the add the respective umbers. There will be as may 2 flag sigals as there are ways of fillig i 2 vacat places i successio by the 5 flags available. By Multiplicatio rule, the umber of ways is 5 4 20. Similarly, there will be as may 3 flag sigals as there are ways of fillig i 3 vacat places i successio by the 5 flags.
138 MATHEMATICS The umber of ways is 5 4 3 60. Cotiuig the same way, we fid that The umber of 4 flag sigals 5 4 3 2 120 ad the umber of 5 flag sigals 5 4 3 2 1 120 Therefore, the required o of sigals 20 + 60 + 120 + 120 320. EXERCISE 7.1 1. How may 3-digit umbers ca be formed from the digits 1, 2, 3, 4 ad 5 assumig that (i) repetitio of the digits is allowed? (ii) repetitio of the digits is ot allowed? 2. How may 3-digit eve umbers ca be formed from the digits 1, 2, 3, 4, 5, 6 if the digits ca be repeated? 3. How may 4-letter code ca be formed usig the first 10 letters of the Eglish alphabet, if o letter ca be repeated? 4. How may 5-digit telephoe umbers ca be costructed usig the digits 0 to 9 if each umber starts with 67 ad o digit appears more tha oce? 5. A coi is tossed 3 times ad the outcomes are recorded. How may possible outcomes are there? 6. Give 5 flags of differet colours, how may differet sigals ca be geerated if each sigal requires the use of 2 flags, oe below the other? 7.3 Permutatios I Example 1 of the previous Sectio, we are actually coutig the differet possible arragemets of the letters such as ROSE, REOS,..., etc. Here, i this list, each arragemet is differet from other. I other words, the order of writig the letters is importat. Each arragemet is called a permutatio of 4 differet letters take all at a time. Now, if we have to determie the umber of 3-letter words, with or without meaig, which ca be formed out of the letters of the word NUMBER, where the repetitio of the letters is ot allowed, we eed to cout the arragemets NUM, NMU, MUN, NUB,..., etc. Here, we are coutig the permutatios of 6 differet letters take 3 at a time. The required umber of words 6 5 4 120 (by usig multiplicatio priciple). If the repetitio of the letters was allowed, the required umber of words would be 6 6 6 216.
PERMUTATIONS AND COMBINATIONS 139 Defiitio 1 A permutatio is a arragemet i a defiite order of a umber of objects take some or all at a time. I the followig sub-sectio, we shall obtai the formula eeded to aswer these questios immediately. 7.3.1 Permutatios whe all the objects are distict Theorem 1 The umber of permutatios of differet objects take r at a time, where 0 < r ad the objects do ot repeat is ( 1) ( 2)...( r + 1), which is deoted by P r. Proof There will be as may permutatios as there are ways of fillig i r vacat places... by r vacat places the objects. The first place ca be filled i ways; followig which, the secod place ca be filled i ( 1) ways, followig which the third place ca be filled i ( 2) ways,..., the rth place ca be filled i ( (r 1)) ways. Therefore, the umber of ways of fillig i r vacat places i successio is ( 1) ( 2)... ( (r 1)) or ( 1) ( 2)... ( r + 1) This expressio for P r is cumbersome ad we eed a otatio which will help to reduce the size of this expressio. The symbol! (read as factorial or factorial ) comes to our rescue. I the followig text we will lear what actually! meas. 7.3.2 Factorial otatio The otatio! represets the product of first atural umbers, i.e., the product 1 2 3... ( 1) is deoted as!. We read this symbol as factorial. Thus, 1 2 3 4... ( 1)! 1 1! 1 2 2! 1 2 3 3! 1 2 3 4 4! ad so o. We defie 0! 1 We ca write 5! 5 4! 5 4 3! 5 4 3 2! 5 4 3 2 1! Clearly, for a atural umber! ( 1)! ( 1) ( 2)! [provided ( 2)] ( 1) ( 2) ( 3)! [provided ( 3)] ad so o.
140 MATHEMATICS Example 5 Evaluate (i) 5! (ii) 7! (iii) 7! 5! Solutio (i) 5! 1 2 3 4 5 120 (ii) 7! 1 2 3 4 5 6 7 5040 ad (iii) 7! 5! 5040 120 4920. Example 6 Compute (i) 7! 5! 12! (ii) ( 10! )(2!) Solutio (i) We have 7! 5! 7 6 5! 5! 7 6 42 ad (ii) 12! ( 10! )( 2! ) ( ) ( 10! ) ( 2) 12 11 10! 6 11 66. Example 7 Evaluate r ( r)!!!, whe 5, r 2. 5! 2! 5 2! Solutio We have to evaluate ( ) (sice 5, r 2) 5! 2! 5 2! We have ( ) 5! 5 4 10 2! 3! 2. Example 8 If 1 1 x +, fid x. 8! 9! 10! Solutio We have 1 1 x + 8! 9 8! 10 9 8! Therefore 1 x 1+ 9 10 9 So x 100. or 10 x 9 10 9 EXERCISE 7.2 1. Evaluate (i) 8! (ii) 4! 3!
PERMUTATIONS AND COMBINATIONS 141 2. Is 3! + 4! 7!? 3. Compute 8! 6! 2! 4. If 1 1 x +, fid x 6! 7! 8!! 5. Evaluate ( r)!, whe (i) 6, r 2 (ii) 9, r 5. 7.3.3 Derivatio of the formula for P r! Pr, 0 r! ( r) Let us ow go back to the stage where we had determied the followig formula: P r ( 1) ( 2)... ( r + 1) Multiplyig umerator ad deomirator by ( r) ( r 1)... 3 2 1, we get ( ) ( ) ( )( )( ) ( r)( r 1)... 3 2 1 1 2... r + 1 r r 1... 3 2 1 P r P! Thus r ( r)!, where 0 < r! ( r ) This is a much more coveiet expressio for P r tha the previous oe.! I particular, whe r, P! 0! Coutig permutatios is merely coutig the umber of ways i which some or all objects at a time are rearraged. Arragig o object at all is the same as leavig behid all the objects ad we kow that there is oly oe way of doig so. Thus, we ca have P 0 1!!! ( 0)!... (1),! Therefore, the formula (1) is applicable for r 0 also.! P, 0 r!. Thus r ( r)
142 MATHEMATICS Theorem 2 The umber of permutatios of differet objects take r at a time, where repetitio is allowed, is r. Proof is very similar to that of Theorem 1 ad is left for the reader to arrive at. Here, we are solvig some of the problems of the pervious Sectio usig the formula for P r to illustrate its usefuless. I Example 1, the required umber of words 4 P 4 4! 24. Here repetitio is ot allowed. If repetitio is allowed, the required umber of words would be 4 4 256. The umber of 3-letter words which ca be formed by the letters of the word NUMBER 6 6! P3 3! 4 5 6 120. Here, i this case also, the repetitio is ot allowed. If the repetitio is allowed,the required umber of words would be 6 3 216. The umber of ways i which a Chairma ad a Vice-Chairma ca be chose from amogst a group of 12 persos assumig that oe perso ca ot hold more tha oe positio, clearly 12 12! P2 11 12 132. 10! 7.3.4 Permutatios whe all the objects are ot distict objects Suppose we have to fid the umber of ways of rearragig the letters of the word ROOT. I this case, the letters of the word are ot all differet. There are 2 Os, which are of the same kid. Let us treat, temporarily, the 2 Os as differet, say, O 1 ad O 2. The umber of permutatios of 4-differet letters, i this case, take all at a time is 4!. Cosider oe of these permutatios say, RO 1 O 2 T. Correspodig to this permutatio,we have 2! permutatios RO 1 O 2 T ad RO 2 O 1 T which will be exactly the same permutatio if O 1 ad O 2 are ot treated as differet, i.e., if O 1 ad O 2 are the same O at both places. Therefore, the required umber of permutatios 4! 2! 3 4 12. Permutatios whe O 1, O 2 are Permutatios whe O 1, O 2 are differet. the same O. RO1O 2T RO2O1T R O O T TO1O2R TO2O1R T O O R
PERMUTATIONS AND COMBINATIONS 143 R O1T O2 R O2T O 1 T O R O 1 2 T O R O 2 1 RTO1O2 RTO2O 1 T R O1O2 T R O2O 1 O 1 O 2 R T O 2 O1 T R O RO T 1 2 O R O T 2 1 O 1 T O 2 R O 2 T O 1 R O 1 R T O2 O 2 R T O 1 O 1 T R O2 O 2 T R O 1 O 1 O2T R O 2 O1 T R R O T O T O R O R T O O T R O O O O R T O R O T O T O R O R T O O T R O O O T R Let us ow fid the umber of ways of rearragig the letters of the word INSTITUTE. I this case there are 9 letters, i which I appears 2 times ad T appears 3 times. Temporarily, let us treat these letters differet ad ame them as I 1, I 2, T 1, T 2, T 3. The umber of permutatios of 9 differet letters, i this case, take all at a time is 9!. Cosider oe such permutatio, say, I 1 NT 1 SI 2 T 2 U E T 3. Here if I 1, I 2 are ot same
144 MATHEMATICS ad T 1, T 2, T 3 are ot same, the I 1, I 2 ca be arraged i 2! ways ad T 1, T 2, T 3 ca be arraged i 3! ways. Therefore, 2! 3! permutatios will be just the same permutatio correspodig to this chose permutatio I 1 NT 1 SI 2 T 2 UET 3. Hece, total umber of differet permutatios will be 9! 2! 3! We ca state (without proof) the followig theorems: Theorem 3 The umber of permutatios of objects, where p objects are of the same kid ad rest are all differet! p!. I fact, we have a more geeral theorem. Theorem 4 The umber of permutatios of objects, where p 1 objects are of oe kid, p 2 are of secod kid,..., p k are of k th kid ad the rest, if ay, are of differet kid is! p! p!... p!. 1 2 k Example 9 Fid the umber of permutatios of the letters of the word ALLAHABAD. Solutio Here, there are 9 objects (letters) of which there are 4A s, 2 L s ad rest are all differet. 9! 5 6 7 8 9 Therefore, the required umber of arragemets 7560 4!2! 2 Example 10 How may 4-digit umbers ca be formed by usig the digits 1 to 9 if repetitio of digits is ot allowed? Solutio Here order matters for example 1234 ad 1324 are two differet umbers. Therefore, there will be as may 4 digit umbers as there are permutatios of 9 differet digits take 4 at a time. 9 9! 9! Therefore, the required 4 digit umbers P 4 9 4! 5! 9 8 7 6 3024. ( ) Example 11 How may umbers lyig betwee 100 ad 1000 ca be formed with the digits 0, 1, 2, 3, 4, 5, if the repetitio of the digits is ot allowed? Solutio Every umber betwee 100 ad 1000 is a 3-digit umber. We, first, have to
PERMUTATIONS AND COMBINATIONS 145 cout the permutatios of 6 digits take 3 at a time. This umber would be 6 P 3. But, these permutatios will iclude those also where 0 is at the 100 s place. For example, 092, 042,..., etc are such umbers which are actually 2-digit umbers ad hece the umber of such umbers has to be subtracted from 6 P 3 to get the required umber. To get the umber of such umbers, we fix 0 at the 100 s place ad rearrage the remaiig 5 digits takig 2 at a time. This umber is 5 P 2. So 6 5 6! 5! The required umber P3 P2 3! 3! 4 5 6 4 5 100 Example 12 Fid the value of such that (i) P5 42 P3,> 4 (ii) 4 1 P 4 P 5, > 4 3 Solutio (i) Give that 5 3 P 42 P or ( 1) ( 2) ( 3) ( 4) 42 ( 1) ( 2) Sice > 4 so ( 1) ( 2) 0 Therefore, by dividig both sides by ( 1) ( 2), we get ( 3 ( 4) 42 or 2 7 30 0 or 2 10 + 3 30 or ( 10) ( + 3) 0 or 10 0 or + 3 0 or 10 or 3 As caot be egative, so 10. (ii) Give that 4 1 P4 P 5 3 Therefore 3 ( 1) ( 2) ( 3) 5( 1) ( 2) ( 3) ( 4) or 3 5 ( 4) [as ( 1) ( 2) ( 3) 0, > 4] or 10.
146 MATHEMATICS Example 13 Fid r, if 5 4 P r 6 5 P r 1. 4 5 Solutio We have 5 P 6 P r r 1 4! 5! 5 6 4! 5 1! or ( r) ( r + ) 5! 6 5! 4! 5 1 5 5 1! or ( r) ( r+ ) ( r)( r ) or (6 r) (5 r) 6 or r 2 11r + 24 0 or r 2 8r 3r + 24 0 or (r 8) (r 3) 0 or r 8 or r 3. Hece r 8, 3. Example 14 Fid the umber of differet 8-letter arragemets that ca be made from the letters of the word DAUGHTER so that (i) all vowels occur together (ii) all vowels do ot occur together. Solutio (i) There are 8 differet letters i the word DAUGHTER, i which there are 3 vowels, amely, A, U ad E. Sice the vowels have to occur together, we ca for the time beig, assume them as a sigle object (AUE). This sigle object together with 5 remaiig letters (objects) will be couted as 6 objects. The we cout permutatios of these 6 objects take all at a time. This umber would be 6 P 6 6!. Correspodig to each of these permutatios, we shall have 3! permutatios of the three vowels A, U, E take all at a time. Hece, by the multiplicatio priciple the required umber of permutatios 6! 3! 4320. (ii) If we have to cout those permutatios i which all vowels are ever together, we first have to fid all possible arragmets of 8 letters take all at a time, which ca be doe i 8! ways. The, we have to subtract from this umber, the umber of permutatios i which the vowels are always together. Therefore, the required umber 8! 6! 3! 6! (7 8 6) 2 6! (28 3) 50 6! 50 720 36000 Example 15 I how may ways ca 4 red, 3 yellow ad 2 gree discs be arraged i a row if the discs of the same colour are idistiguishable? Solutio Total umber of discs are 4 + 3 + 2 9. Out of 9 discs, 4 are of the first kid
PERMUTATIONS AND COMBINATIONS 147 (red), 3 are of the secod kid (yellow) ad 2 are of the third kid (gree). 9! Therefore, the umber of arragemets 1260. 4! 3! 2! Example 16 Fid the umber of arragemets of the letters of the word INDEPENDENCE. I how may of these arragemets, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels ever occur together (iv) do the words begi with I ad ed i P? Solutio There are 12 letters, of which N appears 3 times, E appears 4 times ad D appears 2 times ad the rest are all differet. Therefore 12! The required umber of arragemets 1663200 3! 4! 2! (i) Let us fix P at the extreme left positio, we, the, cout the arragemets of the remaiig 11 letters. Therefore, the required umber of words startig with P (ii) 11! 138600. 3! 2! 4! There are 5 vowels i the give word, which are 4 Es ad 1 I. Sice, they have to always occur together, we treat them as a sigle object EEEEI for the time beig. This sigle object together with 7 remaiig objects will accout for 8 objects. These 8 objects, i which there are 3Ns ad 2 Ds, ca be rearraged i 8! ways. Correspodig to each of these arragemets, the 5 vowels E, E, E, 3! 2! (iii) E ad I ca be rearraged i 5! ways. Therefore, by multiplicatio priciple, 4! the required umber of arragemets 8! 5! 16800 3! 2! 4! The required umber of arragemets the total umber of arragemets (without ay restrictio) the umber of arragemets where all the vowels occur together.
148 MATHEMATICS (iv) 1663200 16800 1646400 Let us fix I ad P at the extreme eds (I at the left ed ad P at the right ed). We are left with 10 letters. Hece, the required umber of arragemets 10! 3!2!4! 12600 EXERCISE 7.3 1. How may 3-digit umbers ca be formed by usig the digits 1 to 9 if o digit is repeated? 2. How may 4-digit umbers are there with o digit repeated? 3. How may 3-digit eve umbers ca be made usig the digits 1, 2, 3, 4, 6, 7, if o digit is repeated? 4. Fid the umber of 4-digit umbers that ca be formed usig the digits 1, 2, 3, 4, 5 if o digit is repeated. How may of these will be eve? 5. From a committee of 8 persos, i how may ways ca we choose a chairma ad a vice chairma assumig oe perso ca ot hold more tha oe positio? 6. Fid if 1 P 3 : P 4 1 : 9. 7. Fid r if (i) 5 P 6 r 2 P (ii) 5 6 r 1 Pr P. r 1 8. How may words, with or without meaig, ca be formed usig all the letters of the word EQUATION, usig each letter exactly oce? 9. How may words, with or without meaig ca be made from the letters of the word MONDAY, assumig that o letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel? 10. I how may of the distict permutatios of the letters i MISSISSIPPI do the four I s ot come together? 11. I how may ways ca the letters of the word PERMUTATIONS be arraged if the (i) words start with P ad ed with S, (ii) vowels are all together, (iii) there are always 4 letters betwee P ad S? 7.4 Combiatios Let us ow assume that there is a group of 3 law teis players X, Y, Z. A team cosistig of 2 players is to be formed. I how may ways ca we do so? Is the team of X ad Y differet from the team of Y ad X? Here, order is ot importat. I fact, there are oly 3 possible ways i which the team could be costructed.
PERMUTATIONS AND COMBINATIONS 149 Fig. 7.3 These are XY, YZ ad ZX (Fig 7.3). Here, each selectio is called a combiatio of 3 differet objects take 2 at a time. I a combiatio, the order is ot importat. Now cosider some more illustratios. Twelve persos meet i a room ad each shakes had with all the others. How do we determie the umber of had shakes. X shakig hads with Y ad Y with X will ot be two differet had shakes. Here, order is ot importat. There will be as may had shakes as there are combiatios of 12 differet thigs take 2 at a time. Seve poits lie o a circle. How may chords ca be draw by joiig these poits pairwise? There will be as may chords as there are combiatios of 7 differet thigs take 2 at a time. Now, we obtai the formula for fidig the umber of combiatios of differet objects take r at a time, deoted by C r.. Suppose we have 4 differet objects A, B, C ad D. Takig 2 at a time, if we have to make combiatios, these will be AB, AC, AD, BC, BD, CD. Here, AB ad BA are the same combiatio as order does ot alter the combiatio. This is why we have ot icluded BA, CA, DA, CB, DB ad DC i this list. There are as may as 6 combiatios of 4 differet objects take 2 at a time, i.e., 4 C 2 6. Correspodig to each combiatio i the list, we ca arrive at 2! permutatios as 2 objects i each combiatio ca be rearraged i 2! ways. Hece, the umber of permutatios 4 C 2 2!. O the other had, the umber of permutatios of 4 differet thigs take 2 at a time 4 P 2. 4! 4 2! 2! Therefore 4 P 2 4 C 2 2! or ( ) Now, let us suppose that we have 5 differet objects A, B, C, D, E. Takig 3 at a time, if we have to make combiatios, these will be ABC, ABD, ABE, BCD, BCE, CDE, ACE, ACD, ADE, BDE. Correspodig to each of these 5 C 3 combiatios, there are 3! permutatios, because, the three objects i each combiatio ca be 4 C 2
150 MATHEMATICS rearraged i 3! ways. Therefore, the total of permutatios 5 C3 3! 5! 5 3! 3! Therefore 5 P 3 5 C 3 3! or ( ) These examples suggest the followig theorem showig relatioship betwee permutaio ad combiatio: Theorem 5 P C!, 0 < r. r r r Proof Correspodig to each combiatio of C r, we have r! permutatios, because r objects i every combiatio ca be rearraged i r! ways. Hece, the total umber of permutatios of differet thigs take r at a time is C r r!. O the other had, it is Pr. Thus P C!, 0 < r. r r r! Remarks 1. From above ( r) I particular, if r, C r r!!! C 1.! 0! 5 C C, i.e., r r ( r) 3!!!. 2. We defie C 0 1, i.e., the umber of combiatios of differet thigs take othig at all is cosidered to be 1. Coutig combiatios is merely coutig the umber of ways i which some or all objects at a time are selected. Selectig othig at all is the same as leavig behid all the objects ad we kow that there is oly oe way of doig so. This way we defie C 0 1.! 3. As 1 C0, the formula 0! 0! Hece ( ) C r! r! r!, 0 r. ( ) C r! is applicable for r 0 also. r! r! ( ) 4.! C r!! ( ) ( r) ( r)! r!! r ( ) C, r
PERMUTATIONS AND COMBINATIONS 151 i.e., selectig r objects out of objects is same as rejectig ( r) objects. 5. C a C b a b or a b, i.e., a + b + 1 Theorem 6 C + C C r r 1 r Proof We have C r!! + Cr 1 + r! r! r 1! r 1! ( ) ( )( + )! r r 1! r! ( )( )! 1! 1! + ( r )( r+ )( r )! 1!! ( r )( r ) 1 1 + r r+ 1! r+ 1+ r ( r 1! )( r )! r ( r+ 1 ) Example 17 If C9 C, fid 8 C17 Solutio We have C9 C8!! 9! 9! 8! 8! i.e., ( ) ( ). ( + 1! ) ( + r) r! 1! + 1 C r or 1 1 or 8 9 or 17 9 8 17 Therefore C C 1. 17 17 Example 18 A committee of 3 persos is to be costituted from a group of 2 me ad 3 wome. I how may ways ca this be doe? How may of these committees would cosist of 1 ma ad 2 wome? Solutio Here, order does ot matter. Therefore, we eed to cout combiatios. There will be as may committees as there are combiatios of 5 differet persos take 3 at a time. Hece, the required umber of ways 5 5! 4 5 C3 10. 3! 2! 2 Now, 1 ma ca be selected from 2 me i 2 C 1 ways ad 2 wome ca be selected from 3 wome i 3 C 2 ways. Therefore, the required umber of committees
152 MATHEMATICS 2! 3! C C 6 1! 1! 2! 1!. 2 3 1 2 Example 19 What is the umber of ways of choosig 4 cards from a pack of 52 playig cards? I how may of these (i) (ii) (iii) (iv) (v) four cards are of the same suit, four cards belog to four differet suits, are face cards, two are red cards ad two are black cards, cards are of the same colour? Solutio There will be as may ways of choosig 4 cards from 52 cards as there are combiatios of 52 differet thigs, take 4 at a time. Therefore The required umber of ways 52 52! 49 50 51 52 C4 4! 48! 2 3 4 270725 (i) There are four suits: diamod, club, spade, heart ad there are 13 cards of each suit. Therefore, there are 13 C 4 ways of choosig 4 diamods. Similarly, there are 13 C 4 ways of choosig 4 clubs, 13 C 4 ways of choosig 4 spades ad 13 C 4 ways of choosig 4 hearts. Therefore The required umber of ways 13 C 4 + 13 C 4 + 13 C 4 + 13 C 4. (ii) (iii) There are13 cards i each suit. 13! 4 2860 4! 9! Therefore, there are 13 C 1 ways of choosig 1 card from 13 cards of diamod, 13 C 1 ways of choosig 1 card from 13 cards of hearts, 13 C 1 ways of choosig 1 card from 13 cards of clubs, 13 C 1 ways of choosig 1 card from 13 cards of spades. Hece, by multiplicatio priciple, the required umber of ways 13 13 13 13 C 1 C 1 C 1 C 1 13 4 There are 12 face cards ad 4 are to be selected out of these 12 cards. This ca be 12! doe i 12 C 4 ways. Therefore, the required umber of ways 495 4! 8!.
PERMUTATIONS AND COMBINATIONS 153 (iv) (v) There are 26 red cards ad 26 black cards. Therefore, the required umber of ways 26 C 2 26 C 2 ( 325) 2 26! 2! 24! 2 105625 4 red cards ca be selected out of 26 red cards i 26 C 4 ways. 4 black cards ca be selected out of 26 black cards i 26 C 4 ways. Therefore, the required umber of ways 26 C 4 + 26 C 4 26! 2 29900. 4! 22! EXERCISE 7.4 1. If C 8 C 2, fid C 2. 2. Determie if (i) 2 C 3 : C 3 12 : 1 (ii) 2 C 3 : C 3 11 : 1 3. How may chords ca be draw through 21 poits o a circle? 4. I how may ways ca a team of 3 boys ad 3 girls be selected from 5 boys ad 4 girls? 5. Fid the umber of ways of selectig 9 balls from 6 red balls, 5 white balls ad 5 blue balls if each selectio cosists of 3 balls of each colour. 6. Determie the umber of 5 card combiatios out of a deck of 52 cards if there is exactly oe ace i each combiatio. 7. I how may ways ca oe select a cricket team of eleve from 17 players i which oly 5 players ca bowl if each cricket team of 11 must iclude exactly 4 bowlers? 8. A bag cotais 5 black ad 6 red balls. Determie the umber of ways i which 2 black ad 3 red balls ca be selected. 9. I how may ways ca a studet choose a programme of 5 courses if 9 courses are available ad 2 specific courses are compulsory for every studet? Miscellaeous Examples Example 20 How may words, with or without meaig, each of 3 vowels ad 2 cosoats ca be formed from the letters of the word INVOLUTE? Solutio I the word INVOLUTE, there are 4 vowels, amely, I,O,E,Uad 4 cosoats, amely, N, V, L ad T.
154 MATHEMATICS The umber of ways of selectig 3 vowels out of 4 4 C 3 4. The umber of ways of selectig 2 cosoats out of 4 4 C 2 6. Therefore, the umber of combiatios of 3 vowels ad 2 cosoats is 4 6 24. Now, each of these 24 combiatios has 5 letters which ca be arraged amog themselves i 5! ways. Therefore, the required umber of differet words is 24 5! 2880. Example 21 A group cosists of 4 girls ad 7 boys. I how may ways ca a team of 5 members be selected if the team has (i) o girl? (ii) at least oe boy ad oe girl? (iii) at least 3 girls? Solutio (i) Sice, the team will ot iclude ay girl, therefore, oly boys are to be selected. 5 boys out of 7 boys ca be selected i 7 C 5 ways. Therefore, the required umber of ways 7 7! 6 7 C5 21 5! 2! 2 (ii) Sice, at least oe boy ad oe girl are to be there i every team. Therefore, the team ca cosist of (a) 1 boy ad 4 girls (b) 2 boys ad 3 girls (c) 3 boys ad 2 girls (d) 4 boys ad 1 girl. 1 boy ad 4 girls ca be selected i 7 C 1 4 C 4 ways. 2 boys ad 3 girls ca be selected i 7 C 2 4 C 3 ways. 3 boys ad 2 girls ca be selected i 7 C 3 4 C 2 ways. 4 boys ad 1 girl ca be selected i 7 C 4 4 C 1 ways. Therefore, the required umber of ways 7 C 1 4 C 4 + 7 C 2 4 C 3 + 7 C 3 4 C 2 + 7 C 4 4 C 1 7 + 84 + 210 + 140 441 (iii) Sice, the team has to cosist of at least 3 girls, the team ca cosist of (a) 3 girls ad 2 boys, or (b) 4 girls ad 1 boy. Note that the team caot have all 5 girls, because, the group has oly 4 girls. 3 girls ad 2 boys ca be selected i 4 C 3 7 C 2 ways. 4 girls ad 1 boy ca be selected i 4 C 4 7 C 1 ways. Therefore, the required umber of ways 4 C 3 7 C 2 + 4 C 4 7 C 1 84 + 7 91
PERMUTATIONS AND COMBINATIONS 155 Example 22 Fid the umber of words with or without meaig which ca be made usig all the letters of the word AGAIN. If these words are writte as i a dictioary, what will be the 50 th word? Solutio There are 5 letters i the word AGAIN, i which A appears 2 times. Therefore, the required umber of words 5! 60 2!. To get the umber of words startig with A, we fix the letter A at the extreme left positio, we the rearrage the remaiig 4 letters take all at a time. There will be as may arragemets of these 4 letters take 4 at a time as there are permutatios of 4 differet thigs take 4 at a time. Hece, the umber of words startig with 4! A 4! 24. The, startig with G, the umber of words 12 as after placig G 2! at the extreme left positio, we are left with the letters A, A, I ad N. Similarly, there are 12 words startig with the ext letter I. Total umber of words so far obtaied 24 + 12 + 12 48. The 49 th word is NAAGI. The 50 th word is NAAIG. Example 23 How may umbers greater tha 1000000 ca be formed by usig the digits 1, 2, 0, 2, 4, 2, 4? Solutio Sice, 1000000 is a 7-digit umber ad the umber of digits to be used is also 7. Therefore, the umbers to be couted will be 7-digit oly. Also, the umbers have to be greater tha 1000000, so they ca begi either with 1, 2 or 4. 6! 4 5 6 The umber of umbers begiig with 1 60, as whe 1 is 3! 2! 2 fixed at the extreme left positio, the remaiig digits to be rearraged will be 0, 2, 2, 2, 4, 4, i which there are 3, 2s ad 2, 4s. Total umbers begiig with 2 6! 3 4 5 6 180 2! 2! 2 ad total umbers begiig with 4 6! 4 5 6 120 3!
156 MATHEMATICS Therefore, the required umber of umbers 60 + 180 + 120 360. Alterative Method 7! The umber of 7-digit arragemets, clearly, 420 3! 2!. But, this will iclude those umbers also, which have 0 at the extreme left positio. The umber of such 6! arragemets 3! 2! (by fixig 0 at the extreme left positio) 60. Therefore, the required umber of umbers 420 60 360. Note If oe or more tha oe digits give i the list is repeated, it will be uderstood that i ay umber, the digits ca be used as may times as is give i the list, e.g., i the above example 1 ad 0 ca be used oly oce whereas 2 ad 4 ca be used 3 times ad 2 times, respectively. Example 24 I how may ways ca 5 girls ad 3 boys be seated i a row so that o two boys are together? Solutio Let us first seat the 5 girls. This ca be doe i 5! ways. For each such arragemet, the three boys ca be seated oly at the cross marked places. G G G G G. There are 6 cross marked places ad the three boys ca be seated i 6 P 3 ways. Hece, by multiplicatio priciple, the total umber of ways 6! 5! 6 P 3 5! 3! 4 5 2 3 4 5 6 14400. Miscellaeous Exercise o Chapter 7 1. How may words, with or without meaig, each of 2 vowels ad 3 cosoats ca be formed from the letters of the word DAUGHTER? 2. How may words, with or without meaig, ca be formed usig all the letters of the word EQUATION at a time so that the vowels ad cosoats occur together? 3. A committee of 7 has to be formed from 9 boys ad 4 girls. I how may ways ca this be doe whe the committee cosists of: (i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls? 4. If the differet permutatios of all the letter of the word EXAMINATION are
PERMUTATIONS AND COMBINATIONS 157 listed as i a dictioary, how may words are there i this list before the first word startig with E? 5. How may 6-digit umbers ca be formed from the digits 0, 1, 3, 5, 7 ad 9 which are divisible by 10 ad o digit is repeated? 6. The Eglish alphabet has 5 vowels ad 21 cosoats. How may words with two differet vowels ad 2 differet cosoats ca be formed from the alphabet? 7. I a examiatio, a questio paper cosists of 12 questios divided ito two parts i.e., Part I ad Part II, cotaiig 5 ad 7 questios, respectively. A studet is required to attempt 8 questios i all, selectig at least 3 from each part. I how may ways ca a studet select the questios? 8. Determie the umber of 5-card combiatios out of a deck of 52 cards if each selectio of 5 cards has exactly oe kig. 9. It is required to seat 5 me ad 4 wome i a row so that the wome occupy the eve places. How may such arragemets are possible? 10. From a class of 25 studets, 10 are to be chose for a excursio party. There are 3 studets who decide that either all of them will joi or oe of them will joi. I how may ways ca the excursio party be chose? 11. I how may ways ca the letters of the word ASSASSINATION be arraged so that all the S s are together? Summary Fudametal priciple of coutig If a evet ca occur i m differet ways, followig which aother evet ca occur i differet ways, the the total umber of occurrece of the evets i the give order is m. The umber of permutatios of differet thigs take r at a time, where! repetitio is ot allowed, is deoted by P r ad is give by P r, ( r)! where 0 r.! 1 2 3...! ( 1)! The umber of permutatios of differet thigs, take r at a time, where repeatitio is allowed, is r. The umber of permutatios of objects take all at a time, where p 1 objects
158 MATHEMATICS are of first kid, p 2 objects are of the secod kid,..., p k objects are of the k th kid ad rest, if ay, are all differet is! p! p!... p!. 1 2 The umber of combiatios of differet thigs take r at a time, deoted by! C r, is give by C r, 0 r. r(! r)! k Historical Note The cocepts of permutatios ad combiatios ca be traced back to the advet of Jaiism i Idia ad perhaps eve earlier. The credit, however, goes to the Jais who treated its subject matter as a self-cotaied topic i mathematics, uder the ame Vikalpa. Amog the Jais, Mahavira, (aroud 850) is perhaps the world s first mathematicia credited with providig the geeral formulae for permutatios ad combiatios. I the 6th cetury B.C., Sushruta, i his medicial work, Sushruta Samhita, asserts that 63 combiatios ca be made out of 6 differet tastes, take oe at a time, two at a time, etc. Pigala, a Saskrit scholar aroud third cetury B.C., gives the method of determiig the umber of combiatios of a give umber of letters, take oe at a time, two at a time, etc. i his work Chhada Sutra. Bhaskaracharya (bor 1114) treated the subject matter of permutatios ad combiatios uder the ame Aka Pasha i his famous work Lilavati. I additio to the geeral formulae for C r ad P r already provided by Mahavira, Bhaskaracharya gives several importat theorems ad results cocerig the subject. Outside Idia, the subject matter of permutatios ad combiatios had its humble begiigs i Chia i the famous book I Kig (Book of chages). It is difficult to give the approximate time of this work, sice i 213 B.C., the emperor had ordered all books ad mauscripts i the coutry to be burt which fortuately was ot completely carried out. Greeks ad later Lati writers also did some scattered work o the theory of permutatios ad combiatios. Some Arabic ad Hebrew writers used the cocepts of permutatios ad combiatios i studyig astroomy. Rabbi be Ezra, for istace, determied the umber of combiatios of kow plaets take two at a time, three at a time ad so o. This was aroud 1140. It appears that Rabbi be Ezra did ot kow
PERMUTATIONS AND COMBINATIONS 159 the formula for C r. However, he was aware that C r C r for specific values ad r. I 1321, Levi Be Gerso, aother Hebrew writer came up with the formulae for P r, P ad the geeral formula for C r. The first book which gives a complete treatmet of the subject matter of permutatios ad combiatios is Ars Cojectadi writte by a Swiss, Jacob Beroulli (1654 1705), posthumously published i 1713. This book cotais essetially the theory of permutatios ad combiatios as is kow today.