PERMUTATIONS AND COMBINATIONS

Transcription

1 5. PERMUTATIONS AND COMBINATIONS 1. INTRODUCTION The mai subject of this chapter is coutig. Give a set of objects the problem is to arrage some or all of them accordig to some order or to select some or all of them accordig to some specificatio. 2. FUNDAMENTAL PRINCIPLE OF COUNTING The rule of sum: If a first task ca be performed i m ways, while a secod task ca be performed i ways, ad the tasks caot be performed simultaeously, the performig either oe of these tasks ca be accomplished i ay oe of total m+ ways. The rule of product: lf a procedure ca be broke dow ito first ad secod stages, ad if there are m possible outcomes for the first stage ad if, for each of these outcomes, there are possible outcomes for the secod stage, the the total procedure ca be carried out, i the desigated order, i a total of m ways. Illustratio 1: There are three statios, A. B ad C. Five routes for goig from statio A to statio B ad four routes for goig from statio B to statio C. Fid the umber of differet ways through which a perso ca go from A to C via B. Sol: This problem is a applicatio of the Fudametal Priciple of Coutig. The rule of product ca be used to solve this questio easily. Give there are five routes for goig from A to B ad four routes for goig from B to C. A B C Figure 5.1 Hece, by the fudametal priciple of multiplicatio the total umber of differet ways = 5 x 4 (i.e., A to B ad the B to C) = 20 ways Illustratio 2: A hall has 12 gates. I how may ways, ca a ma eter the hall through oe gate ad come out through a differet gate. Sol: The rule of product ca be used to solve this problem.

2 5.2 Permutatios ad Combiatios There are 12 ways of eterig the hall. After eterig the hall the ma ca come out through ay of 11 differet gates. Hece, by the fudametal priciple of multiplicatio, the total umber of ways are 12 x 11 = 132 ways. Illustratio 3: How may umbers betwee 10 ad 10,000 ca be formed by usig the digits 1, 2, 3, 4, 5 if (i) o digit is repeated i ay umber. (ii) digits ca be repeated. Sol: The umbers betwee 10 ad 10,000 ca be either two digit, three digit or four digit umbers. We cosider each of these cases ad try to fid the umber of possibilities usig 1, 2, 3, 4 ad 5. Fially, we add them up to get the desired result. (i) Number of two digit umbers = 5 x 4 = 20 Number of three digit umbers = 5 x 4 x 3 = 60 Number of four digit umbers = 5 x 4 x 3 x 2 = 120 Total umber of umbers = = 200 (ii) Number of two digit umbers = 5 x 5 = 25 Number of three digit umbers = 5 x 5 x 5 = 125 Number of four digit umbers = = 625 Total umber of umbers = = FACTORIAL NOTATION A efficiet way of writig a product of several cosecutive itegers is the factorial otatio. The umber! (read as -factorial ) is defied as follows : For ay positive iteger ;! = ( 1)( 2).. (3)(2)(1); For istace, 4! = = 24 Note: (i)! = ( 1)( 2) ;! =.( 1)! ; 0! = 1! = 1; (2)! = 2.![ (2 1)] (ii)! = ( 1)! = ( 1)( 2)! = ( 1)( 2) ( 3)!! (iii) ( 1).. ( r + 1) = ( r)! Illustratio 4: Fid the sum of terms of the series whose th term is!. (JEE ADVANCED) Sol: Represet the geeral term i this questio as a differece of two terms ad the add them up to fid the aswer. The required sum = (1)! + 2(2)! + 3(3)! +. + (!) = (2 1)! + (3 1)2! + (4 1)3! [( + 1) 1]! = (2! 1!) + (3! 2!) + (4! 3!) [( + 1)!!] = ( + 1)! 1 4. PERMUTATION Each of the differet arragemets, which ca be made by takig some or all of a umber of objects is called permutatio. The umber of permutatios of differet objects take r at a time is represeted as! P r = = ( 1)( 2) ( r + 1) (where, 0 r ) ( r)! Note: (i) I permutatio, the order of the items plays a importat role. (ii) The umber of all permutatios of distict objects take all at a time is!

3 Mathematics 5.3 Illustratio 5: If 56 P r+6 : 54 P r+3 = : 1, fid r P 2 Sol: Use the formula for P r. We have, 56 r Pr+ 3 P = = 1 41 P 2 = 41 x 40 = ! (51 r)! = (50 r)! 54! 1 56 x 55 (51 r) = r = 41 Illustratio 6: Three me have 4 coats, 5 waist coats ad 6 caps. I how may ways ca they wear them? (JEE ADVANCED) Sol: Use the cocept ad uderstadig of Permutatio, i.e. arragemet to fid the aswer. The total umber of ways i which three me ca wear 4 coats is the umber of arragemets of 4 differet coats take 3 at a time. So, three me ca wear 4 coats i 4 P 3 ways. Similarly, 5 waist coats ad 6 caps ca be wor by three me i 5 P 3 ad 6 P 3 ways respectively. Hece, the required o. of ways = 4 P 3 5 P 3 6 P 3 = (4!) (5 4 3) (6 5 4) = Illustratio 7: Suppose 8 people eter a evet i a swim meet. I how may ways could the gold, silver, ad broze prizes be awarded? (JEE ADVANCED) Sol: Use the formula for P r. The required umber of ways is a arragemet of 3 people out of 8 i.e. 8 8! P 3 = = = = ! MASTERJEE CONCEPTS The followig two steps are ivolved i the solutio of a permutatio problem: Step 1: Recogizig the objects ad the places ivolved i the problem. Step 2: Checkig whether the repetitio of the objects is allowed or ot. Uday Kira G (JEE 2012, AIR 102) 4.1 Permutatio with Repetitio These kids of problems occur with permutatios of differet objects i which some of the objects ca be repeated. The o. of permutatios of differet objects take r at a time whe each object may be repeated ay umber of times is r. Illustratio 8: A studet appears i a objective test which cotais 10 multiple choice questios. Each questio has four choices i which oe is the correct optio. What maximum umber of differet aswers ca the studet give? How will the aswer chage if each questio may have more tha oe correct aswers? (JEE ADVANCED) Sol: Use the cocept of Permutatio with Repetitio. For the first part each questio has four possible aswers. So, the total possible aswers = times = For the secod part. Suppose the choices for each questio are deoted by A, B, C ad D. Now the choice A is either correct or icorrect (two ways) similarly the other choices are either correct or icorrect. Thus, this particular questio ca have = 16 possible aswers. But this icludes the case whe all the four choices are icorrect. Thus the total umber of aswers = 15. Now, as there are 10 questios ad each questio has 15 possible aswers. Therefore the total umber of aswers =

4 5.4 Permutatios ad Combiatios MASTERJEE CONCEPTS The m or m dilemma Let us start with a example. Q. There are 7 letters ad 5 letter-boxes. I how may ways ca you put the letters i the boxes? Sol: This is the typically cofusig questio asked frequetly from the P & C area. Let s see how you ca solve this type of questio. The Exhaustive Approach: Oe way to solve this questio is through (what we will call) The Exhaustive Approach. While solvig such problems, first decide which of the items (letters ad letter-boxes here) is exhaustive. Exhaustive here meas the etity which is sure to be used up completely. I this example, all the letters are sure to be placed i the boxes, whereas there is o such costrait as regards the letter-boxes. Some boxes could go empty. Havig decided this, just go by the optios we have for all the istaces of the exhaustive item ad you have your aswer. As you ca see here, every letter has 5 boxes to choose from. Thus the total would be (5 x 5 x...7 times) = (5 7 ) A similar questio could be: I how may ways ca 10 rigs be wor o 5 figers? Try it yourself. Chimay S Puradare (JEE 2012, AIR 698) 4.2 Permutatio of Alike Objects This kid of problems ivolve permutatios of differet objects i which some of them are similar. The umber of permutatios of objects take all at a time i which, p are alike objects of oe kid, q are alike objects of secod kid & r are alike objects of a third kid ad the rest ( (p + q + r)) are all differet is! p!q!r! Illustratio 9: Determie the umber of permutatios of the letters of the word SIMPLETION take all at a time. Sol: I the give word the letter I occurs twice ad the remaiig letters occur oly oce. So, the cocept of Permutatio of Alike Objects is used to fid out the aswer. There are 10 letters i the word SIMPLETION ad out of these 10 letters two are idetical. So, just selectig all 10 objects at a time will give twice the actual result. Hece, the umber of permutatios of takig all the letters at a time = 10 P 10 / 2! = 10! / 2! = Permutatio uder Restrictio (a) The umber of permutatios of differet objects, take r at a time, whe a particular object is to be always icluded i each arragemet, is r. 1 P r 1 The umber of permutatios of differet objects, take r at a time, whe a particular object is ever take i each arragemet is 1 P r. (b) Strig method: The umber of permutatios of differet objects, take all at a time, whe m specified objects always come together is m! ( m + 1)!. (c) Gap Method: The umber of permutatios whe o two give objects occur together. I order to fid the umber of permutatios whe o two give objects occur together.

5 Mathematics 5.5 (a) First of all, put the m objects for which there is o restrictio, i a lie. These m objects ca be arraged i m! ways. (b) The cout the umber of gaps betwee every two m objects for which there is o restrictio, icludig the ed positios. Number of such gaps will be (m + 1). (c) If m is the umber of objects for which there is o restrictio ad is the umber of objects, two of which are ot allowed to occur together, the the required umber of ways = m! m+1 C! The umber of permutatios whe two types of objects are to be arraged alterately (a) If their umbers differ by 1 put the object whose umber is greater i the first, third, fifth... places, etc. ad the other object i the secod, fourth, sixth... places. (b) If the umber of two types of objects is same, cosider two cases separately keepig the first type of object i the first, third, fifth places, etc. ad the secod type of object i the first, third, fifth places... ad the add. 4.4 No-Cosecutive Selectio The umber of selectios of r cosecutive objects out of objects i a row = - r + 1, whe r < The umber of selectios of r cosecutive objects out of objects alog a circle = 1, whe r = Illustratio 10: Fid the umbers betwee 300 ad 3000 that ca be formed with the digits 0, 1, 2, 3, 4 ad 5, where o digit is repeated i ay umber. Sol: The umbers betwee 300 ad 3000 ca either be a three digit umber or a four digit umber. The solutio is divided ito these two differet cases ad their sum will give us the desired result. Ay umber betwee 300 ad 3000 must be of three or four digits. Case I: Whe umber is of three digits: The hudreds place ca be filled by ay oe of the three digits 3, 4 ad 5 i 3 ways. The remaiig two places ca be filled by the remaiig five digits i 5 P 2, ways. The umber of umbers formed i this case = 3 5 P 2 = 3 5! 3! =60 Case II: Whe umber is of four digits: The thousads place ca be filled by ay oe of the two digits 1 ad 2 i 2 ways ad the remaiig three places ca be filled by the remaiig five digits i 5 P 3 ways. The umber of umbers formed i this case = 2 5 P 3 = 2 5! 2! = 120 Total umbers = = 180 Illustratio 11: How may words ca be formed from the letters of the word ARTICLE, so that vowels occupy the eve places? Sol: Clearly, this is a example of Permutatio uder Restrictio. We idetify the eve places ad the odd places ad try to fid the umber of ways i which the vowels ad cosoats ca fill the spaces. There are seve places: 3 eve ad 4 odd i which we have to fill 3 vowels ad 4 cosoats. The umber of words = 3 P 3. 4 P 4 = 3! 4! = 6 24 = 144. Illustratio 12: How may differet words ca be formed with the letters of the word ORDINATE so that (a) Four vowels occupy the odd places (b) Begiig with O (c) Begiig with O ad edig with E. Sol: The cocept of Permutatio uder Restrictio ca be used to solve this problem. There are 4 vowels ad 4 cosoats. Total 8 letters. (a) No. of words = 4! 4! = = 576. Because 4 vowels are to be adjusted i 4 odd place ad the 4 cosoats i the remaiig 4 eve places.

6 5.6 Permutatios ad Combiatios (b) 7! ways, O beig fixed. (c) 6! ways, O fixed i first ad E fixed i last. Illustratio 13: Fid the umber of ways i which 5 boys ad 5 girls ca be seated i a row so that (a) No two girls may sit together. (b) All the girls sit together ad all the boys sit together (c) All the girls are ever together. (JEE ADVANCED) Sol: Sice the umber of girls ad the umber of boys are equal they have to sit alterately. This ca be used to solve (a). For (b), we keep the girls together ad arrage the boys i five places. Also, the girls ca be arraged amogst themselves i 5! ways. This gives us the umber of arragemets. Use the aswer of the secod part to fid (c). (a) 5 boys ca be seated i a row i 5 P 5 = 5! ways. Now, i the 6 gaps betwee 5 boys, the 5 girls ca be arraged i 6 P 5 ways. Hece, the umber of ways i which o two girls sit together = 5! 5 P 5 = 5! 6! (b) The two groups of girls ad boys ca be arraged i 2! ways. 5 girls ca be arraged amog themselves i 5! ways. Similarly, 5 boys ca be arraged amog themselves i 5! ways. Hece, by the fudametal priciple of coutig, the total umber of requisite seatig arragemets = 2!(5! 5!) = 2(5!) 2. (c) The total umber of ways i which all the girls are ever together = Total umber of arragemets Total umber of arragemets i which all the girls are always together = 10! 5! 6! Illustratio 14: The letters of the word OUGHT are writte i all possible orders ad these words are writte out as i a dictioary. Fid the rak of the word TOUGH i this dictioary. Sol: The word TOUGH will appear after all the words that start with G, H ad O. The we look at the secod letter of the words startig with T ad the third. Hece, the rak of the word TOUGH will be oe more tha the sum of all the possibilities just metioed. Total umber of letters i the word OUGHT is 5 ad all the five letters are differet, the alphabetical order of these letters is G, H, O, T, U. Number of words begiig with G = 4! = 24 Number of words begiig with H = 4! = 24 Number of words begiig with O = 4! = 24 Number of words begiig with TG = 3! = 6 Number of words begiig with TH = 3! = 6 Number of words begiig with TOG = 2! = 2 Number of words begiig with TOH = 2! = 2 Next come the words begiig with TOU ad TOUH is the first word begiig with TOU. Rak of TOUGH i the dictioary = = 89 Illustratio 15: There are 21 balls which are either white or black ad the balls of same color are alike. Fid the umber of white balls so that, the umber of arragemets of these balls i a row is maximum. (JEE ADVANCED) Sol: The property of a biomial coefficiet ca be used to solve this questio. Let there be r white balls so that the umber of arragemets of these balls i a row be maximum. Number of arragemets of these balls is 21! A = A will be maximum whe r = or 21 1 i.e. 10 or 11 r!(21 r)! 2 2

7 Mathematics 5.7 Illustratio 16: The umber plates of cars must cotai 3 letters of the alphabet deotig the place ad area to which its ower belogs. This is to be followed by a three-digit umber. How may differet umber plates ca be formed if: (i) Repetitio of letters ad digits is ot allowed. (ii) Repetitio of letters ad digits is allowed. (JEE ADVANCED) Sol: This is a simple applicatio of Permutatio with ad without repetitio. There are 26 letters of alphabet ad 10 digits from 0 to 9. (i) Whe repetitio is ot allowed 3 letters selected i ways 3 digit umbers are = 9 x 9 x 8 (as zero ca t be i the hudreds place) The Number of plates = 26 x 25 x 24 x 9 x 9 x 8 = (ii) Whe repetitio is allowed 3 letters are selected ways 3 digit umbers are = = 900 The umber of plates = 26 x 26 x 26 x 900 = MASTERJEE CONCEPTS Costrait based arragemet Let us start with a example first: Q. I how may ways ca the word VARIETY be arraged so that exactly 2 vowels are together? The problem could be easier if oe or all vowels were to be kept together. Is t it? Well, we will do exactly that! Wheever questio ivolvig costraits that has choices (2 vowels could be IE or AI or AE) are asked, go for the backward approach. Rather tha fidig the favorable cases, subtract the ufavorable oes from the total possible cases. This method is more reliable. So, the solutio to the above questio would be: (Total arragemets of VARIETY) (Arragemets with o vowels together + Arragemets with all the vowels together) Vaibhav Krisha (JEE 2009, AIR 22) 5. COMBINATION Each of the differet groups or selectio which ca be made by some or all of a umber of give objects without referece to the order of the objects i each group is called a combiatio.! The umber of all combiatios of objects, take r at a time is geerally deoted by C(, r) or C r = p r!( r)! (0 r ) = r r! Note: (a) The umber of ways of selectig r objects out of objects, is the same as the umber of ways i which the remaiig ( - r) ca be selected ad rejected. (b) The combiatio otatio also represets the biomial coefficiet. That is, the biomial coefficiet C r is the combiatio of elemets chose r at a time.

8 5.8 Permutatios ad Combiatios (c) (a) C r = C r (b) C 1 r C + + r 1 = C r (c) C x = C y x = y or x + y = (d) If is eve, the the greatest value of C r is C /2 C + 2 (e) If is odd, the the greatest value of C r is 1 (f) C 0 + C r C = 2 (g) C + +1 C + +2 C C = 2 C +1 (d) Compariso of permutatio ad combiatio Permutatios Differet orderigs or arragemet of the r objects are differet permutatios! P r = ( r)! Clue words: arragemet, schedule, order Combiatios Each choice or subset of r object give oe combiatio. Order withi the group of r objects does ot matter.! C r = ( r)!r! Clue words: group, committee, sample, selectio, subset. Illustratio 17: A basketball coach must select two attackers ad two defeders from amog three attackers ad five defeders. How may differet combiatios of attackers ad defeders ca he select? Sol: The umber of ways two attackers ad two defeders ca be selected is 3 C 2 ad 5 C 2 respectively. He ca select i 3 C 2 5 C 2 = = 30 differet combiatios. Illustratio 18: A soccer team of 11 players is to be chose from 30 boys, of whom 4 ca play oly i goal, 12 ca play oly as forwards ad the remaiig 14 i ay of the other positios. If the team is to iclude five forwards ad of course, oe goalkeeper, i how may ways ca it be made up? Sol: Proceed accordig to the previous questio. There are: 4 C 1 ways of choosig the goalkeeper. l2 C 5 ways of choosig the forwards ad 14 C 5 ways of choosig the other 5 players. That is, 4 C 1 12 C 5 14 C 5 combiatios altogether = 4 x 792 x 2002 = Combiatios uder Restrictios (a) Number of ways of choosig r objects out of differet objects if p particular objects must be excluded. Cosider the objects A 1, A 2, A 3,...,A p, A p+1,..., A. If the p objects A 1, A 2...A p are to be excluded the we will have to select r objects from the remaiig p objects (A p+1, A p+2,.. A ). Hece the required umber of ways = ( p) C r (b) Number of ways of choosig r objects out of differet objects if p particular objects must be icluded (p r). Cosider the objects A 1, A 2, A 3... A p, A p+1,. A. If the p particular objects A 1, A 2,... A p (say) must be icluded i the selectio the to complete the selectio, we must select (r p) more objects to complete the selectio. These objects are to be selected from the remaiig p objects. Hece, the required umber of ways = p C r p (c) The total umber of combiatios of differet objects take oe or more at a time = 2-1.

9 Mathematics Combiatios of Alike Objects (a) The umber of combiatios of idetical objects takig (r ) at a time is 1. (b) The umber of ways of selectig r objects out of idetical objects is + 1. (c) If out of (p + q + r + s) objects, p are alike of oe kid, q are alike of a secod kid, r are alike of the third kid ad s are differet, the the total umber of combiatios is (p + 1)(q + 1)(r + 1)2 s 1 Note: The list of alike objects ca be exteded further 4. The umber of ways i which r objects ca be selected from a group of objects of which p are idetical, is t t p p C r, if r p ad C r if r > p 0 r= p Illustratio 19: There are 4 orages, 5 apples ad 6 magoes i a fruit basket. I how may ways ca a perso select fruits from amog the fruits i the basket? Sol: Use the cocept of Combiatio of Alike objects described above. Here, we cosider all fruits of the same type as idetical. Zero or more orages ca be selected out of 4 idetical orages i = 5 ways. Zero or more apples ca be selected out of 5 idetical apples i = 6 ways. Zero or more magoes ca be selected out of 6 idetical magoes i = 7 ways The total umber of selectios whe all the three types of fruits are selected (the umber of ay type of fruit may be zero) = = 210. But i oe of these selectios umber of each type of fruit is zero ad hece there is o selectio, this must be excluded. The required umber = = 209. Cautio: Whe all fruits of same type are differet, the umber of selectios = ( 4 C C C 4 )( 5 C C C 5 ) ( 6 C C 6 ) 1 = 2 4 x 2 5 x = Illustratio 20: How may four digit umbers are there whose decimal otatio cotais ot more tha two distict digits? Sol: A four digit umber ca cosist of either oly oe digit or two digits as per the questio. Clearly, there are ie four digit umbers with the same digit. Similarly, calculate the umber of four digit umbers with two distict digits ad hece the sum gives us the desired result. Evidetly ay umber so formed of four digits cotais (i) Oly oe digit (like 1111, 2222,...) ad there are 9 umbers. (ii) Two digits (a) if zero is oe of the two, the the oe more ca be ayoe of the ie, ad these two digits ca be arraged i 9 C 1 [ 3 C C C C 3 ] = 63. (b) if zero is ot oe of them, the two of the digits have to be selected from 9, ad these i 9 C 2 [ 4 C C C 3 ] = 504 Hece, the total umber of required umbers = 567. two ca be arraged Illustratio 21: I how may ways ca a cricket team of eleve players be chose out of a batch of 15 players, if (a) there is o restrictio o the selectio (b) a particular player is always chose (c) a particular player is ever chose

10 5.10 Permutatios ad Combiatios Sol: Usig the cocept of combiatio of alike objects we ca get the aswer (a) The total umber of ways of selectig 11 players out of 15 is = 15 C 11 = 15 C = 15 C 4 = = (b) A particular player is always chose. This meas that 10 players are selected out of the remaiig 14 players. The required umber of ways = 14 C 10 = 14 C = 14 C 4 = 1001 (c) The umber of ways = 14 C 11 = 14 C 3 = = Illustratio 22: I a plae there are 37 straight lies, of which 13 pass through the poit A ad 11 pass through the poit B. Moreover, o three lies pass through oe poit, o lie passes both poits A ad B. ad o two are parallel. Fid the umber of poits of itersectio of the straight lies. Sol: Two o parallel straight lies give us a poit of itersectio. Usig this idea we fid the total umber of poits of itersectio. Care should be take that a poit is ot couted more tha oce. The umber of poits of itersectio of 37 straight lies is 37 C 2. But 13 straight lies out of the give 37 straight lies pass through the same poit A. Therefore, istead of gettig 13 C 2 poits, we get merely oe poit A. Similarly, 11 straight lies out of the give 37 straight lies itersect at poit B. Therefore, istead of gettig 11 C 2 poits, we get oly poit B. Hece, the umber of itersectio poits of the lies i 37 C 2 13 C 2 11 C = 535. Illustratio 23: How may five-digit umbers ca be made havig exactly two idetical digits? Sol: Note that zero caot occupy the first place. So we divide the solutio ito two cases whe the commo digit is 0 ad otherwise. Calculate the umber of possibilities i these two cases ad their sum gives us the desired result. Case I: Two idetical digits are 0, 0. The umber of ways to select three more digits is 9 C 3. The umber of arragemets of these five digits is ( 5! 2! ) 4! = = 36. Hece, the umber of such umbers is 9 C 3 36 = 3024 (i) Case II: Two idetical digits are (1, 1) or (2, 2) or... or (9, 9). If 0 is icluded, the umber of ways of selectio of two more digits is 8 C 2. The umber of ways of arragemets of these five digits is 5! / 2! -4! / 2! =48. Therefore, the umber of such umbers is 8 C If 0 is ot icluded, the selectio of three more digits is 8 C 3. Therefore, the umber of such umbers is 8 C 3 5! / 2! = 8 C Hece, the total umber of five-digit umbers with idetical digits (1.1)...(9.9) is 9 ( 8 C C 3 60) = (ii) From Eqs. (i) ad (ii), the required umber of umbers is = Illustratio 24: How may words ca be made with letters of the word INTERMEDIATE if (i) The words either begi with I or ed with E, (ii) The vowels ad cosoats alterate i the words, (iii) The vowels are always cosecutive, (iv) The relative order of vowels ad cosoats does ot chage, (v) No vowel is i betwee two cosoats, (vi) The order of vowels does ot chage?

11 Mathematics 5.11 Sol: This is a applicatio of Permutatio uder restrictio ad Permutatio of Alike objects. Proceed accordig to the give coditios. (i) The required umber of words = (the umber of words without restrictio) (the umber of words begiig with I) (the umber of word edig with E) + (the umber of words begiig with I ad edig with E) (because words begiig with I as well as words edig with E cotai some words begiig with I ad edig with E). 12! The umber of words without restrictio = 2! 3! 2! ( There are 12 letters i which there are two I s, three E s ad two T s). The umber of words begiig with I = 11! 2!3! ( With E i the extreme left place we are left to arrage 11 letters INTERMEDIATE i which there are two T s ad three E s). The umber of words edig with E = 11! 2!2!2! ( With E i the extreme right place we are left to arrage 11 letters INTERMEDIATE i which there are two I s. two E s ad two T s.) The umber of words begiig with I ad edig with E = 10! 2!2! ( With I i the extreme left ad E i the extreme right places we are left to arrage 10 letters INTERMEDIATE i the which there are two T s ad Two E s). 12! 11! 11! 10! 10! the required umber or words = + = 83 10! ( ) = 2!3!2! 2!3! 2!2!2! 2!2! 2!3!2! 24 (ii) There are 6 vowels ad 6 cosoats. So, the umber of words i which vowels ad cosoats alterate = (the umber of words i which vowels occupy odd places ad cosoats occupy eve places) + (the umber of words i which cosoats occupy odd places ad vowels occupy eve places) = 6! 6! 6! 6! + = 2. 2!3! 2! 2! 2!3! 6! 6!. 2!3! 2! = (iii) Cosiderig the 6 vowels IEEIAE as oe object, the umber of arragemets of this with 6 cosoats = 7! 2! ( there are two T s i the cosoats). For each of these arragemets, the 6 cosecutive vowels ca be arraged amog themselves i 7! 6! The required umber of words = (as above) = ! 2!3! (iv) The relative order of vowels ad cosoats will ot chage if i the arragemets of letters, the vowels occupy places of vowels, i.e.. 1 st, 4 th, 7 th, 9 th, 10 th, 12 th places ad cosoats occupy their places, i.e., 2d. 3rd. 5th. 6th. 8th. 11th places, the required umber of words 6! 6! = !3! 2! (v) No vowel will be betwee two cosoats if all the cosoats become cosecutive the required umber of words = the umber of arragemets whe all the cosoats are cosecutive 7! 6! = (as above) = !3! 2! (vi) The order of vowels will ot chage if o two vowels iterchage places, i.e., i the arragemet all the vowels are treated as idetical. (For example LATE. ATLE. TLAE. etc.. have the same order of vowels A. E. But LETA, ETLA. TLEA. etc.. have chaged order of vowels A. E. So. LATE is couted but LETA is ot. If A, E, are take as idetical say V the LVTV does ot give a ew arragemet by iterchagig V, V. 6! 2!3!.

12 5.12 Permutatios ad Combiatios The required umber of words, = The umber of arragemets of 12 letters i which 6 vowels are treated as idetical = 12! 6!2! ( there are two T s also). Illustratio 25: Idia ad South Africa play a oe day iteratioal series util oe team wis 4 matches. No match eds i a draw. Fid, i how may ways ca the series ca be wo. (JEE ADVANCED) Sol: The team who wis the series is the team with more umber of wis. The losig team wis either 0 or 1 or 2 or 3 matches. Usig this we fid the umber of ways i which a team ca wi. Let I for Idia ad S for South Africa. We ca arrage I ad S to show the wis for Idia ad South Africa respectively Suppose Idia wis the series, the the last match is always wo by Idia. Wis of S Wis of I No. of ways (i) (ii) 1 4 4! / 3! = 4 (iii) 2 4 5! 2!3! =10 (iv) 3 4 Total o. of ways = 35 6! 3!3! = 20 I the same umber of ways South Africa ca wi the series Total o. of ways i which the series ca be wo = 35 2 = 70 Illustratio 26: There are p itermediate railway statios o a railway lie from oe termial to aother. I how may ways ca a trai stop at three of these itermediate statios, if o two of these statios (where it stops) are to be cosecutive? (JEE ADVANCED) Sol: The trai stops oly at three itermediate statios implies that the trai does ot stop at (p 3) statios. Usig this idea we proceed further to get the aswer. The problem the reduces to the followig: I how may ways ca three objects be placed amog (p 3) objects i a row such that o two of them are ext to each other (at most 1 object is to be placed betwee ay two of these (p 3) objects). Sice there are (p 2) positios to place the three objects, the required umber of ways = p 2 C 3. Illustratio 27: Five balls are to be placed i three boxes. Each ca hold all the five balls. I how may differet ways ca we place the balls so that o box remais empty if (i) balls ad boxes are all differet (ii) balls are idetical but boxes are differet (iii) balls are differet but boxes are idetical (iv) balls as well as boxes are idetical (v) balls as well as boxes are idetical but boxes are kept i a row? Sol: Use the differet cases of combiatio to solve the questio accordig to the give coditios As o box is to remai empty, boxes ca have balls i the followig umbers: Possibilities 1, 1, 3 or 1, 2, 2

13 Mathematics 5.13 (i) The umber of ways to distribute the balls i groups of 1, 1, 3 = 5 C 1 4 C 1 3 C 3. But the boxes ca iterchage their cotet, o exchage gives a ew way whe boxes cotaiig balls i equal umber iterchage. the total umber of ways to distribute 1, 1, 3 balls to the boxes = 5 C 1 4 C 1 3 C 3 3! 2! Similarly, the total umber of ways to distribute 1, 2, 2 balls to the boxes = 5 C 1 4 C 2 2 C 2 3! 2! the required umber of ways = 5 C 1 4 C 1 3 C 3 3! 2! + 5 C 1 4 C 1 2 C 2 3! = = = 150 2! Note: Writig the whole aswer i tabular form. Possibilities Combiatios Permutatios 1, 1, 3 5 C 1 4 C 1 3 C 3 5 C 1 4 C 1 3 C 3 3! 2! = = 60 1, 2, 2 5 C 1 4 C 2 2 C 5 2 C 1 4 C 2 2 C 2 3! 2! = = 90 the required umber of ways = = 150. (ii) Whe balls are idetical but boxes are differet the umber of combiatios will be 1 i each case. the required umber of ways = 1 3! 3! +1 2! 2! = = 6 (iii) Whe the balls are differet ad boxes are idetical, the umber of arragemets will be 1 i each case. the required umber of ways = 5 C 1 4 C 1 3 C C 1 4 C 2 2 C 2 = = = 50 (iv) Whe balls as well as boxes are idetical, the umber of combiatios ad arragemets will be 1 each i both cases. the required umber of ways = = 2 (v) Whe boxes are kept i a row, they will be treated as differet. So, i this case the umber of ways will be the same as i (ii). Illustratio 28: There are m poits o oe straight lie AB ad poits o aother straight lie AC, oe of them beig A. How may triagles ca be formed with these poits as vertices? How may ca be formed if poit A is also icluded? Sol: A triagle has three vertices, so we select two poits o oe lie ad oe o the other ad vice versa. Also, cosider the case whe oe poit of the triagle is the itersectio of the two lies. To get a triagle, we either take two poits o AB ad oe poit o AC. or oe poit o AB ad two poits o AC. Therefore, the umber of triagles, we obtai m(m 1) ( 1) = ( m C 2 )( C 1 ) + ( m C 1 )( C 2 ) = + m = m(m 1 + 1) = 1 m(m + 2) 2 If the poit A is icluded, we get m additioal triagles. Thus, i this case we get = m 2 m(m + ) (m + 2) + m = Divisio ito Groups triagles. (a) The umber of ways i which (m + ) differet objects ca be divided ito two uequal groups cotaiig m ad objects respectively is (m + )!. m!! If m =, the groups are equal ad i this case the umber of divisio is (2)! ; as it is possible to iterchage the!!2! two groups without obtaiig a ew distributio.

14 5.14 Permutatios ad Combiatios (b) However, if 2 objects are to be divided equally betwee two persos the the umber of ways (2)! (2)! = 2! =!!2!!! (c) The umber of ways i which (m + + p) differet objects ca be divided ito three uequal groups cotaiig (m + + p) m, ad p objects respectively is =, m p m!!p! (3)! If m = = P the the umber of groups =. However, if 3 objects are to be divided equally amog three!!!3! (3)! persos the the umber of ways =!!!3! 3! = (3)! 3 (!) 15! For example, the umber of ways i which 15 recruits ca be divided ito three equal groups is ad the 5!5!5!3! 15! umber of ways i which they ca be drafted ito three differet regimets, five i each, is 5!5!5! (d) The umber of ways i which m differet objects ca be divided equally ito m groups if order of groups is m! ot importat is m (!) m! (e) The umber of ways i which m differet objects ca be divided equally ito m groups if the order of groups m! (m)! is importat is m! = m m (!) m! (!) Illustratio 29: I how may ways ca 12 balls be divided betwee 2 boys, oe receivig 5 ad the other 7 balls? Sol: Simple applicatio of divisio of objects ito groups. Sice the order is importat, the umber of ways i which 12 differet balls ca be divided betwee two boys who each get 5 ad 7 balls respectively, is 12! ! 2! = 2 = !7! ( )7! Alterative: The first boy ca be give 5 balls out of 12 balls i 12 C 5 ways. The secod boy ca be give the remaiig 7 balls i oe way. But the order is importat (the boys ca iterchage 2 ways). Thus, the required umber of ways = 12 C 5 1 2! = 12! 5!7! 2 = !.2 = ! Illustratio 30: Fid the umber of ways i which 9 differet toys ca be distributed amog 4 childre belogig to differet age groups i such a way that the distributio amog the 3 elder childre is eve ad the yougest oe is to receive oe toy more. (JEE ADVANCED) Sol: Usig the cocept of divisio of objects ito groups we ca solve this problem very easily. The distributio should be 2, 2, 2 ad 3 to the yougest. Now, 3 toys for the yougest ca be selected i 9 C 3 ways, the remaiig 6 toys ca be divided ito three equal groups i 6! ways ad ca be distributed i 3! ways. 3 (2!).3! 6! 9! Thus, the required umber of ways = 9 C 3. 3! = 3 3 (2!) 3! 3!(2!)

15 Mathematics 5.15 Illustratio 31: Divide 50 objects i 5 groups of size 10, 10, 10, 15 ad 5 objects. Also fid the umber of distributios? 50! Sol: Same as the above questio. Number of ways of dividig 50 objects ito 5 groups as give = 3 (10!) (15)!(5)!(3)! 50! Number of ways of distributig 50 objects ito above formed groups = 3 (10)!.(15)!(5)!3! 5! MASTERJEE CONCEPTS Idetical Objects ad Distict Choices Questios ivolvig idetical objects ted to be tricky, especially whe the choices that they have are distict. May choices image 1 A example would be : Q. I how may ways ca we place 10 idetical orages i 3 distict baskets, such that every basket has at least 2 orages each? Oe method is to place 2 orages i every basket ad make cases for the rest of them. However, i such questios, the other approach results i fewer cases, ad hece, simpler calculatios ad a more efficiet solutio For this questio: Divide 10 ito groups of 3 rather tha placig 2 i each ad dividig the remaiig four. Vaibhav Gupta (JEE 2009, AIR 54) 6. CIRCULAR PERMUTATION Let us cosider that persos A, B, C ad D are sittig aroud a roud table. If all of them (A, B, C, D) are shifted oe place i a aticlockwise order, the we will get fig.(b) from fig.(a). Now, if we shift A, B, C, D i aticlockwise order agai, we will get fig. (c). We shift them oce more ad we will get fig.(d); ad i the ext time fig(a). C B A D D B C A B D A C A D C B A B C D D A B C C D A B B C D A a b c d Figure 5.2 Thus, we see that if 4 persos are sittig at a roud table, they ca be shifted four times ad the four differet arragemets thus obtaied will be same, because the aticlockwise order of A, B, C, D does ot chage. But if A, B, C ad D are sittig i a row ad they are shifted i such a order that the last occupies the place of first, the the four arragemets will be differet. Thus, if there are 4 objects, the for each circular arragemet umber of liear arragemets is 4. Similarly, if differet objects are arraged alog a circle, for each circular arragemet the umber of liear arragemets is. Therefore, the umber of circular arragemets of differet objects = the umber of liear arragemets of differet objects / =!/() = ( 1)!

16 5.16 Permutatios ad Combiatios Clockwise ad Aticlockwise Arragemets Let the four persos A, B, C ad D sit at a roud table i aticlockwise as well as clockwise directios. These two arragemets are differet. But if four flowers R (red), G (gree), Y (yellow) ad B (blue) are arraged to form a garlad i aticlockwise ad i clockwise order, the the two arragemets are same because if we view the garlad from oe side the four flowers R, G, Y, B will appear i aticlockwise directio ad if see from the other side the four flowers will appear i the clockwise directio. Here the two arragemets will be cosidered as oe arragemet because the order of flowers does ot chage, rather oly the side of observatio chages. Here, two permutatios will be couted as oe. C C Y Y D B B D B G G B A A R R Figure 5.3 Therefore, whe clockwise ad aticlockwise arragemets are ot differet, i.e. whe observatios ca be made from both sides, the umber of circular arragemets of differet objects is ( 1)!/2 Cosider five persos A, B, C, D, E o the circumferece of a circular table i a order which has o head. Now, shiftig A, B, C, D, E oe positio i aticlockwise directio we will get arragemets as follows: We see, that arragemets i all figures are differet. The umber of circular permutatio of differet objects take all at a time is ( 1)!, if clockwise ad aticlockwise orders are take as differet. Note: (a) The umber of circular permutatios of differet objects take r at a time P r /r, whe clockwise ad aticlockwise orders are treated as differet. P r /2r, whe clockwise ad aticlockwise orders are treated as same. (b) The umber of circular permutatios of differet objects altogether P / = ( 1)!, whe clockwise ad aticlockwise order are treated as differet, P /2 = 1/2( 1)!, whe the above two orders are treated as same. Illustratio 32: I how may ways ca 5 Idias ad 4 Eglishme be seated at a roud table if (a) There is o restrictio, (b) All the four Eglishme sit together, (c) All four Eglishme do t sit together, (d) No two Eglishme sit together. Sol: Clearly, this is a case of Circular Permutatio. Usig the formula ( 1)!, we ca fid the aswer accordig to the give cases. (a) Total umber of persos = = 9. These 9 persos ca be seated at the roud table i 8! Ways. Required umber of ways = 8! (b) Regardig 4 Eglishme as oe perso, we have oly i.e. 6 persos. (c) These 6 persos ca be seated at the roud table i 5! ways. Also, the 4 Eglishme ca be arraged amog themselves i 4! ways. the required umber of ways = 5! 4! The total umber of arragemets whe there is o restrictio = 8!; the umber of arragemets whe all the four Eglish me sit together = 5! 4! The umber of arragemets whe all the four Eglishme do t sit together = 8! 5! 4!

17 Mathematics 5.17 (d) As there is o restrictio o Idias, we first arrage the 5 Idias. Now, 5 Idias ca be seated aroud a table i 4! ways. If a Eglishma sits betwee two Idias, the o two Eglishme will sit together. Now, there are 5 places for 4 Eglish me, therefore, 4 Eglishme ca be seated i 5 P 4 ways. The required umber of ways = 4! 5 P 4 = 4 5! Illustratio 33: Cosider 21 differet pearls o a ecklace. How may ways ca the pearls be placed i this ecklace such that 3 specific pearls always remais together? Sol: This is the case of circular permutatio whe there is o distictio betwee clockwise ad aticlockwise arragemets. After fixig the places of three pearls. Treatig 3 specific pearls = 1 uit. So we have ow 18 pearls + 1 uit = 19 ad the umber of arragemets will be (19 1)! = 18!. Also the umber of ways 3 pearls ca be arraged betwee themselves is 3! = 6. As there is o distictio betwee the clockwise ad aticlockwise arragemets, the required umber of arragemets = 1 18!. 6 = 3 (18!). 2 Illustratio 34: Six persos A, B, C, D, E ad F are to be seated at a circular table. Fid the umber of ways this ca be doe if A must have either B or C o his right ad B must have either C or D o his right. Sol: Fix the positio of some of the persos relative to each other as per the questio ad arrage the remaiig i the seats available. Whe A has B or C to his right we have either AB or AC Whe B has C or D to his right we have BC or BD. Thus, we must have ABC or ABD or AC ad BD. For ABC, D, E, F i a circular umber of ways = 3! = 6 For ABD, C, E, F i a circular umber of ways = 3! = 6 For AC, BD E, F the umber of ways = 3! = 6 Hece, the required umber of ways = LINEAR EQUATIONS WITH UNIT COEFFICIENTS Cosider the equatio x 1 + x 2 + x x k = m, i k variables whose sum must always be m, The umber of o-egative solutios to the above equatio is give by the fictitious partitio method which is stated as: Method of fictitious partitio: Number of ways i which idetical objects may be distributed amog p persos if each perso may receive oe, oe or more objects is = +p 1 C. Coefficiet Method (a) The umber of o-egative itegral solutios of equatio x 1 + x x r = = The umber of ways of distributig idetical objects amog r persos whe each perso ca get zero, oe or more objects = coeff. of x i [(1 + x + x x )(1 + x + x x ) (1 + x + x x ) upto r factors] = coeff. of x i (1 + x + x x ) r r x = coeff. of x i 1 x = coeff. of x i (1 x +1 ) r (1 x) r = coeff. of x i (1 x) r [leavig terms cotaiig powers of x greater tha ] = +r 1 C r 1

18 5.18 Permutatios ad Combiatios Note: If is a positive iteger, the (1 x) = 1 + ( ) (x) + 1! ( )( 1) 2! ( x) 2 + ( )( 1)( 2) 3! ( x) 3 + to = 1 + ( + 1) ( + 1)( + 2) x + x 2 + x 3 +. To = 1 + C 1! 2! 3! 1 x + +1 C 2 x C 3 x 3 + to Coeff. X r i (1 x) = +r 1 C coeff. of x i (1 x) r = +r 1 C = +r 1 C r 1 (b) The umber of positive itegral solutios for the equatio x 1 + x x r = = The umber of ways of distributig idetical objects amog r persos whe each perso ca get at least oe object = coeff. of x i [x + x x ) (x + x x ) (x +x x ). Upto r factors] = coeff. of x i (x + x x ) r = coeff. of x r 1 x i x 1 x = coeff. Of x r i (1 x ) r (1 x) r = coeff. Of x r i (1 x) r [Leavig terms cotaiig powers of x greater tha r] = r+r 1 C r 1 = 1 C r 1. r Illustratio 35: How may itegral solutios are there to x + y + z + w = 29, whe x 1, y 2, z 3 ad w 0? (JEE ADVANCED) Sol: Applicatio of multiomial theorem. x + y + z + w = 29 x 1, y 2, z 3, w 0 x 1 0, y 2 0, z 3 0, w 0 Let x 1 = x 1, x 2 = y 2, x 3 = z 3 x = x 1 + 1, y = x 2 + 2, z = x ad the x 1 0, x 2 0, x 3 0, w 0 From (i), x x x w = 29 x 1 + x 2 + x 3 + w = 23 Hece, the total umber of solutios = C 4 1 = 26 C 3 = (i) Illustratio 36: Fid the umber of o-egative itegral solutio 3x + y + z = 24. Sol: Applicatio of multiomial theorem. 3x + y + z = 24, x 0, y 0, z 0 Let x = k y + z = 24 3k Here, k 24. Hece, 0 k 8 The total umber of itegral solutios of (1) is 24 3k+2 1 C 2 1 = 25 3k C 1 = 25 3k Hece, the total umber of solutios of the origial equatio = (25 3k) = k = = k= 0 k= 0 k= 0 Illustratio 37: Fid the umber of solutios of the equatio x + y + z = 6, where x, y, z W.. (i) Sol: The umber of solutios = C 3 1 = 8 C 2 = 280.

19 Mathematics 5.19 Illustratio 38: How may itegers are there betwee 1 ad havig the sum of the digits as 18? (JEE ADVANCED) Sol: Let the digits be a 1,, a 6 ad use multiomial theorem we get the aswer. Ay umber betwee 1 ad must be of less tha seve digits. Therefore, it must be of the form a 1 a 2 a 3 a 4 a 5 a 6 Where a 1. a 2.a 3.a 4.a 5.a 6 {0, 1, 2, 9} Thus a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 18, where 0 a i 9, i = 1, 2, 9 The required umber of ways = coeff. of x 18 i (1 + x + x 2 + x 9 ) x = coeff. of x 18 i 1 x 6 = coeff. of x 18 i [(1 x 10 ) 6 (1 x) 6 ] ; = coeff. of x 18 i [(1 6 C 1 x 10 ) (1 x) 6 ] [leavig terms cotaiig powers of x greater tha 18] = coeff. of x 18 i (1 x) 6 6 C 1. coeff. of x 8 i (1 x) 6 = C C 5 = 23 C C 5 = = = MASTERJEE CONCEPTS m differet white balls ad differet red balls are to be arraged i a lie such that the balls of the same colour are always together = (m!! 2!) m differet white balls ad differet red balls are to be arraged i a lie such that all the red balls are together = ((m + 1)!!) m differet white balls ad differet red balls are to be arraged i a lie such that o two red balls are together (m 1) = ( m+1 C m!!) m differet white balls ad m differet red balls are to be arraged i a lie such that colour of the balls is alteratig = (2 (m!) 2 ) m idetical white balls ad differet red balls are to be arraged i a lie such that o two red balls are together (m 1) = ( m+1 C ) m idetical white balls ad differet red balls are to be arraged i a lie such that o two red balls are together (m 1) = ( m+1 C!) If objects are arraged i a lie the umber of selectios of r objects ( 2r 1) such that o two objects are adjacet is same umber of ways of arragig r idetical white balls ad r idetical red ball i a lie such that o two balls are together = ( r+1 C r ). e.g. suppose there are statios o trais s route ad a trai has to stop at r statios such that o two statios are adjacet. The umber of ways must be r+1 C r. suppose there are N seats i a particular row of a theatre. The umber of ways of makig people sit (N 2 1) such that o two people sit side by side is same as umber of ways of arragig N idetical white balls (empty seats) ad differet red balls ( people) such that o two red balls are together. The required umber of ways are N +1 C!. Nitish Jhawar (JEE 2009, AIR 7)

20 5.20 Permutatios ad Combiatios 8. DIVISIBILITY OF NUMBERS The followig table shows the coditios of divisibility of some umbers Divisible by Coditio 2 Whose last digit is eve 3 sum of whose digits is divisible by 3 4 whose last two digits umber is divisible by 4 5 whose last digit is either 0 or 5 6 which is divisible by both 2 ad 3 7 If you double the last digit ad subtract it from the rest of the umber, aswer is a multiple of 7 8 whose last three digits umber is divisible by 8 9 sum of whose digits is divisible by 9 10 Whose last digit is 0 11 If you sum every secod digit ad the subtract sum of all other digits, aswer is a multiple of whose last two digits are divisible by 25 Illustratio 39: How may four digit umbers ca be made with the digits 0, 1, 2, 3, 4, 5 which are divisible by 3 (digits beig urepeated i the same umber)? How may of these will be divisible by 6? (JEE ADVANCED) Sol: A umber is divisible by 3 if the sum of the digits is divisible by 3. This reduces the problem to the umber of o-egative itegral solutios of equatio x 1 + x x r =. Here, = 15; so two digits are to be omitted whose sum is 3 or 6 or 9. Hece, the umber of four digits ca be made by either 1, 2, 4, 5 or 0, 3, 4, 5 (omittig two digits whose sum is 3) 0, 1, 3, 5 or 0, 2, 3, 4 (omittig two digits whose sum 6) 0, 1, 2, 3 (omittig two digits whose sum is 9) The umber of 4-digit umbers that ca be made with 1, 2, 4, 5 = 4 P 4 = 4! The umber of 4-digit umbers that ca be made by the digits i ay oe of remaiig four groups (each cotaiig 0) = 4! 3! The required umber of 4-digit umbers divisible by 3 = 4! + 4(4! 3!) = (24 6) = 96 Now, a umber is divisible by 6 if it is eve as well as divisible by 3. So, the umber of 4-digit umbers divisible by 6 that ca be made with 1, 2, 4, 5 = 2 3! ( the umber should have a eve digit i the uits places). The umber of umbers of 4 digits, divisible by 6, that ca be made with 0, 3, 4, 5 = (3! 2!) + 3! ( The umber should have 4 or 0 i uits place ad 0 should ot come i thousads place). Similarly, the umber of umbers of 4 digits, divisible by 6, that ca be made with 0, 1, 2, 3 = (3! 2!) + 3! The umber of 4-digit umbers divisible by 6 that ca be made with the digits 0, 1, 3 = 3! The umber of umbers of 4 digits, divisible by 6, that ca be made with 0, 2, 3, 4 = (3! 2!) + (3! 2!) + 3! ( The umber should have 4 or 2 or 0 i uits place ad 0 should ot come i thousads place)

21 Mathematics 5.21 the required 4-digit umbers divisible by 6 = 2 3! + (3! 2!) + (3! 2!) + 3! + 3! + (3! 2!) + (3! 2!) + (3! 2!) + 3! = = SUM OF NUMBERS (a) For give differet digits a 1, a 2, a 3 a the sum of the digits i the uits place of all umbers formed (if umbers are ot repeated) is (a 1 + a 2 + a a ) ( 1)! i.e. (sum of the digits) ( 1)! (b) Sum of the total umbers which ca be formed with give differet digits a 1, a 2, a 3 a is (a + a 2 + a a ) ( 1)! ( times) Illustratio 40: Fid the sum of all 4 digit umbers formed usig the digits 1, 2, 4 ad 6. Sol: Use formula, Sum = (a 1 + a 2 + a a ) ( 1)! (111. N times) Usig formula, Sum = ( ) 3! (1111) = = Alterate: Here, the total 4-digit umbers will be 4! = 24. So, every digit will occur 6 times at every oe of the four places. Sice the sum of the give digits = = 13. So, the sum of all the digits at every place of all the 24 umbers = 13 6 = 78. The sum of the values of all the digits At first place = 78 At the tes place = 780 At the hudreds place = 7800 At the thousads place = The required sum = FACTORS OF NATURAL NUMBERS Let N = p a. q b. r c where p, q, r are distict primes & a, b, c are atural umbers, the: (a) The total umber of divisors of N icludig 1 ad N are = (a + 1) (b + 1) (c + 1) (b) The sum of these divisors is = (p 0 + p 1 + p p a ) (q 0 + q 1 + q q b ) (r 0 + r 1 + r r c ) (c) The umber of ways i which N ca be resolved as a product of two factors is 1 (a 1+ 1)(a 2+ 1)( a 3+ 1)... if N is ot a perfect square = 2 1 (a 1+ 1)(a 2+ 1)(a 3+ 1) if N is a perfect square 2 (d) The umber of ways i which a composite umber N ca be resolved ito two factors which are relatively prime (or coprime) to each other is equal to 2 1 where is the umber of differet prime factors i N Illustratio 41: Fid the umber of factors of the umber (excludig 1 ad the umber itself). Fid also the sum of these divisors. Sol: Factorise ito its product of primes ad the use the cocept of combiatio to fid the aswer =