STUDY PACKAGE. Subject : Mathematics Topic : Permutation and Combination Available Online :

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1 fo/u fopkjr Hkh# tu] ugha vkjehks dke] foifr s[k NksM+s rqjar e/;e eu dj ';kea iq#"k flag ladyi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /kez iz.ksrk l~xq# Jh j.knksm+klth egkjkt STUDY PACKAGE Subject : Mathematics Topic : Permutatio ad Combiatio Available Olie : R Idex 1. Theory 2. Short Revisio 3. Exercise (Ex = 6) 4. Assertio & Reaso 5. Que. from Compt. Exams Yrs. Que. from IIT-JEE(Advaced) Yrs. Que. from AIEEE (JEE Mai) Studet s Name : Class Roll No. : : Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bod Classes, Zoe-2, M.P. NAGAR, Bhopal : , , WhatsApp

2 Permutatio ad Combiatio Permutatios are arragemets ad combiatios are selectios. I this chapter we discuss the methods of coutig of arragemets ad selectios. The basic results ad formulas are as follows: 1. Fudametal Priciple of Coutig : ( i ) Priciple of Multiplicatio: If a evet ca occur i m differet ways, followig which aother evet ca occur i differet ways, the total umber of differet ways of simultaeous occurrece of both the evets i a defiite order is m. ( ii) Priciple of Additio: If a evet ca occur i m differet ways, ad aother evet ca occur i differet ways, the exactly oe of the evets ca happe i m + ways. Example # 1 There are 8 buses ruig from Kota to Jaipur ad 10 buses ruig from Jaipur to Delhi. I how may ways a perso ca travel from Kota to Delhi via Jaipur by bus. Solutio. Let E 1 be the evet of travellig from Kota to Jaipur & E 2 be the evet of travellig from Jaipur to Delhi by the perso. E 1 ca happe i 8 ways ad E 2 ca happe i 10 ways. Sice both the evets E 1 ad E 2 are to be happeed i order, simultaeously, the umber of ways = 8 10 = 80. Example # 2 How may umbers betwee 10 ad 10,000 ca be formed by usig the digits 1, 2, 3, 4, 5 if (i) No digit is repeated i ay umber. (ii) Digits ca be repeated. Solutio. (i) Number of two digit umbers = 5 4 = 20 Number of three digit umbers = = 60 Number of four digit umbers = = 120 Total = 200 (ii) Number of two digit umbers = 5 5 = 25 Number of three digit umbers = = 125 Number of four digit umbers = = 625 Total = How may 4 digit umbers are there, without repetitio of digits, if each umber is divisible by 5. As Usig 6 differet flags, how may differet sigals ca be made by usig atleast three flags, arragig oe above the other. As Arragemet : If P r deotes the umber of permutatios of differet thigs, takig r at a time, the! P r = ( 1) ( 2)... ( r + 1) = ( r)! NOTE : (i) factorials of egative itegers are ot defied. (ii) 0! = 1! = 1 ; (iii) P =! =. ( 1)! (iv) (2)! = 2.! [ (2 1)] Example # 3: How may umbers of three digits ca be formed usig the digits 1, 2, 3, 4, 5, without repetitio of digits. How may of these are eve. Solutio.: Three places are to be filled with 5 differet objects. Number of ways = 5 P 3 = = 60 For the 2d part, uit digit ca be filled i two ways & the remaiig two digits ca be filled i 4 P 2 ways. Number of eve umbers = 2 4 P 2 = 24. Example # 4: If all the letters of the word 'QUEST' are arraged i all possible ways ad put i dictioary order, the fid the rak of the give word. Solutio.: Number of words begiig with E = 4 P 4 = 24 Number of wards begiig with QE = 3 P 3 = 6 Number of words begiig with QS = 6 Number of words begiig withqt = 6. Next word is 'QUEST' its rak is = Fid the sum of all four digit umbers (without repetitio of digits) formed usig the digits 1, 2, 3, 4, 5. As Fid '', if 1 P 3 : P 4 = 1 : 9. As Six horses take part i a race. I how may ways ca these horses come i the first, secod ad third place, if a particular horse is amog the three wiers (Assume No Ties). As Circular Permutatio : The umber of circular permutatios of differet thigs take all at a time is; ( 1)!. If clockwise & ati clockwise circular permutatios are cosidered to be same, ( 1)! the it is. 2 Note: Number of circular permutatios of thigs whe p alike ad the rest differet take all at a time ( 1)! distiguishig clockwise ad aticlockwise arragemet is. p! Example # 5: I how may ways ca we arrage 6 differet flowers i a circle. I how may ways we ca form a garlad usig these flowers. Solutio.: The umber of circular arragemets of 6 differet flowers = (6 1)! = 120 Whe we form a garlad, clockwise ad aticlockwise arragemets are similar. Therefore, the umber of ways of formig garlad = 2 1 (6 1)! = 60. Example # 6: I how may ways 6 persos ca sit at a roud table, if two of them prefer to sit together. Solutio.: Let P 1, P 3, P 4, P 5, P 6 be the persos, where P 1 wat to sit together. Regard these perso as 5 objects. They ca be arraged i a circle i (5 1)! = 24. Now P 1 P 2 ca be arraged i 2! ways. Thus the total umber of ways = 24 2 = I how may ways the letters of the word 'MONDAY' ca be writte aroud a circle if the vowels are to be separated i ay arragemet. As I how may ways we ca form a garlad usig 3 differet red flowers, 5 differet yellow flowers ad 4 differet blue flowers, if flowers of same colour must be together. As Selectio : If C r deotes the umber of combiatios of differet thigs take r at a time, the! P C r = = r where r ; N ad r W. r! ( r)! r! NOTE : (i) C r = C r (ii) C r + C r 1 = + 1 C r (iii) C r = 0 if r {0, 1, 2, 3..., } Example # 7 Fiftee players are selected for a cricket match. Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phoe : , page 2 of 20

3 (i) I how may ways the playig 11 ca be selected (ii) I how may ways the playig 11 ca be selected icludig a particular player. (iii) I how may ways the playig 11 ca be selected excludig two particular players. Solutio. (i) 11 players are to be selected from 15 Number of ways = 15 C 11 = (ii) Sice oe player is already icluded, we have to select 10 from the remaiig 14 Number of ways = 14 C 10 = (iii) Sice two players are to be excluded, we have to select 11 from the remaiig 13. Number of ways = 13 C 11 = 78. Example # 8 If 49 C 3r 2 = 49 C 2r + 1, fid 'r'. Solutio. C r = C s if either r = s or r + s =. Thus 3r 2 = 2r + 1 r = 3 or 3r 2 + 2r + 1 = 49 5r 1 = 49 r = 10 r = 3, 10 Example # 9 A regular polygo has 20 sides. How may triagles ca be draw by usig the vertices, but ot usig the sides. Solutio. The first vertex ca be selected i 20 ways. The remaiig two are to be selected from 17 vertices so that they are ot cosecutive. This ca be doe i 17 C 2 16 ways. The total umber of ways = 20 ( 17 C 2 16) But i this method, each selectio is repeated thrice ( C2 16) Number of triagles = = Example # persos are sittig i a row. I how may ways we ca select three of them if adjacet persos are ot selected. Solutio. Let P 1, P 3, P 4, P 5, P 6, P 7, P 8, P 9, P 10 be the persos sittig i this order. If three are selected (o cosecutive) the 7 are left out. Let PPPPPPP be the left out & q, q, q be the selected. The umber of ways i which these 3 q's ca be placed ito the 8 positios betwee the P's (icludig extremes) is the umber ways of required selectio. Thus umber of ways = 8 C 3 = 56. Example # 11 I how may ways we ca select 4 letters from the letters of the word M SS SS PP. Solutio. M SSSS PP Number of ways of selectig 4 alike letters = 2 C 1 = 2. Number of ways of selectig 3 alike ad 1 differet letters = 2 C 1 3 C 1 = 6 Number of ways of selectig 2 alike ad 2 alike letters = 3 C 2 = 3 Number of ways of selectig 2 alike & 2 differet = 3 C 1 3 C 2 = 9 Number of ways of selectig 4 differet = 4 C 4 = 1 Total = I how may ways 7 persos ca be selected from amog 5 Idia, 4 British & 2 Chiese, if atleast two are to be selected from each coutry. As poits lie i a plae, of which 4 poits are colliear. Barrig these 4 poits o three of the 10 poits are colliear. How may quadrilaterals ca be draw. As I how may ways 5 boys & 5 girls ca sit at a roud table so that girls & boys sit alterate. As I how may ways 4 persos ca occupy 10 chairs i a row, if o two sit o adjacet chairs. As I how may ways we ca select 3 letters of the word PROPORTION. As The umber of permutatios of '' thigs, take all at a time, whe 'p' of them are similar & of oe type, q of them are similar & of aother type, 'r' of them are similar & of a third type & the remaiig! (p + q + r) are all differet is. p! q! r! Example # 12 I how may ways we ca arrage 3 red flowers, 4 yellow flowers ad 5 white flowers i a row. I how may ways this is possible if the white flowers are to be separated i ay arragemet (Flowers of same colour are idetical). Solutio. Total we have 12 flowers 3 red, 4 yellow ad 5 white. 12! Number of arragemets = = ! 4! 5! For the secod part, first arrage 3 red & 4 yellow 7! This ca be doe i = 35 ways 3! 4! Now select 5 places from amog 8 places (icludig extremes) & put the white flowers there. This ca be doe i 8 C 5 = 56. The umber of ways for the 2 d part = = Example # 13 I how may ways the letters of the word "ARRANGE" ca be arraged without alterig the relative positios of vowels & cosoats. 4! Solutio. The cosoats i their positios ca be arraged i = 12 ways. 2! 3! The vowels i their positios ca be arraged i = 3 ways 2! Total umber of arragemets = 12 3 = How may words ca be formed usig the letters of the word ASSESSMENT if each word begi with A ad ed with T. As If all the letters of the word ARRANGE are arraged i all possible ways, i how may of words we will have the A's ot together ad also the R's ot together. As How may arragemets ca be made by takig four letters of the word MISSISSIPPI. As Formatio of Groups : Number of ways i which (m + + p) differet thigs ca be divided ito three m p! differet groups cotaiig m, & p thigs respectively is, m!! p! Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phoe : , page 3 of 20

4 (3)! If m = = p ad the groups have idetical qualitative characteristic the the umber of groups =.!!! 3! (3)! However, if 3 thigs are to be divided equally amog three people the the umber of ways = 3.! Ex differet toys are to be distributed to three childre equally. I how may ways this ca be doe. Solutio. The problem is to divide 12 differet thigs ito three differet groups. 12! Number of ways = = ! 4! 4! Example # 15 I how may ways 10 persos ca be divided ito 5 pairs. Solutio. We have each group havig 2 persos ad the qualitative characteristic are same (Sice there is o purpose metioed or ames for each pair). 10! Thus the umber of ways = 5 = 945. (2!) 5! persos eter a lift from groud floor of a buildig which stops i 10 floors (excludig groud floor). If is kow that persos will leave the lift i groups of 2, 3, & 4 i differet floors. I how may ways this ca happe. As ! 17. I how may ways oe ca make four equal heaps usig a pack of 52 playig cards.as. 4 (13!) 4! 18. I how may ways 11 differet books ca be parcelled ito four packets so that three of the packets cotai 11! 3 books each ad oe of 2 books, if all packets have the same destiatio. As. 4 (3!) 2 7. Selectio of oe or more objects (a) Number of ways i which atleast oe object be selected out of '' distict objects is C 1 + C 2 + C C = 2 1 (b) Number of ways i which atleast oe object may be selected out of 'p' alike objects of oe type 'q' alike objects of secod type ad 'r' alike of third type is (p + 1) (q + 1) (r + 1) 1 (c) Number of ways i which atleast oe object may be selected from '' objects where 'p' alike of oe type 'q' alike of secod type ad 'r' alike of third type ad rest (p + q + r) are differet, is (p + 1) (q + 1) (r + 1) 2 (p + q + r) 1 Example # 16 There are 12 differet books o a shelf. I how may ways we ca select atleast oe of them. Solutio. We may select 1 book, 2 books,..., 12 books. The umber of ways = 12 C C C 12 = = 4095 Example # 17 There are 12 fruits i a basket of which 5 are apples, 4 magoes ad 3 baaas (fruits of same species are idetical). How may ways are there to select atleast oe fruit. Solutio. Let x be the umber of apples beig selected y be the umber of magoes beig selected ad z be the umber of baaas beig selected. The x = 0, 1, 2, 3, 4, 5 y = 0, 1, 2, 3, 4 z = 0, 1, 2, 3 Total umber of triplets (x, y, z) is = 120 Exclude (0, 0, 0) Number of combiatios = = 119. Self Practice Problems 19. I a shelf there are 5 physics, 4 chemistry ad 3 mathematics books. How may combiatios are there if (i) books of same subject are differet (ii) books of same subject are idetical. As. (i) 4095 (ii) From 5 apples, 4 magoes & 3 baaas i how may ways we ca select atleast two fruits of each variety if (i) fruits of same species are idetical (ii) fruits of same species are differet. As. (i) 24 (ii) Multiomial Theorem: Coefficiet of x r i expasio of (1 x) = +r 1 C r ( N) Number of ways i which it is possible to make a selectio from m + + p = N thigs, where p are alike of oe kid, m alike of secod kid & alike of third kid take r at a time is give by coefficiet of x r i the expasio of (1 + x + x x p ) (1 + x + x x m ) (1 + x + x x ). (i) For example the umber of ways i which a selectio of four letters ca be made from the letters of the word PROPORTION is give by coefficiet of x 4 i (1 + x + x 2 + x 3 ) (1 + x + x 2 ) (1 + x + x 2 ) (1 + x) (1 + x) (1 + x). (ii) Method of fictious partitio : Number of ways i which idetical thigs may be distributed amog p persos if each perso may receive oe, oe or more thigs is; +p 1 C. Example # 18: Fid the umber of solutios of the equatio x + y + z = 6, where x, y, z W. Solutio. Number of solutios = coefficiet of x 6 i (1 + x + x x 6 ) 3 = coefficiet of x 6 i (1 x 7 ) 3 (1 x) 3 = coefficiet of x 6 i (1 x) = = 8 C 6 6 = 28. Example # 19: I a bakery four types of biscuits are available. I how may ways a perso ca buy 10 biscuits if he decide to take atleast oe biscuit of each variety. Solutio. Let x be the umber of biscuits the perso select from first variety, y from the secod, z from the third ad w from the fourth variety. The the umber of ways = umber of solutios of the equatio x + y + z + w = 10. where x = 1, 2,...,7 y = 1, 2,...,7 z = 1, 2,...,7 w = 1, 2,...,7 Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phoe : , page 4 of 20

5 This is equal to = coefficiet of x 10 i (x + x x 7 ) 4 = coefficiet of x 6 i (1 + x x 6 ) 4 = coefficiet of x 6 i (1 x 7 ) 4 (1 x) = coefficiet x 6 i (1 x) 4 = = Self Practice Problems: 21. Three distiguishable dice are rolled. I how may ways we ca get a total 15. As I how may ways we ca give 5 apples, 4 magoes ad 3 orages (fruits of same species are similar) to three persos if each may receive oe, oe or more. As 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 umbers of divisors of N icludig 1 & N is = (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) Number of ways i which N ca be resolved as a product of two factors is 1 (a 1)(b 1)(c 1)... if N is ot a perfect square = 1 2 (a 1)(b 1)(c 1)... 1 if N is a perfect square 2 (d) Number 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. Example # 20 Fid the umber of divisors of Also fid the sum of all divisors. Solutio = Number of divisors = (1+ 1) (3 + 1) (2 + 1) = 24 sum of divisors = (1 + 2) ( ) ( ) = Example # 21 I how may ways 8100 ca be resolved ito product of two factors. Solutio = Number of ways = 2 1 ((2 + 1) (4 + 1) (2 + 1) + 1) = How may divisors of 9000 are eve but ot divisible by 4. Also fid the sum of all such divisors. As. 12, I how may ways the umber 8100 ca be writte as product of two coprime factors. As Let there be '' types of objects, with each type cotaiig atleast r objects. The the umber of ways of arragig r objects i a row is r. Example # 22 How may 3 digit umbers ca be formed by usig the digits 0, 1, 2, 3, 4, 5. I how may of these we have atleast oe digit repeated. Solutio. We have to fill three places usig 6 objects (repeatatio allowed), 0 caot be at 100 th place. The umber of umbers = 180. Number of umbers i which o digit is repeated = 100 Number of umbers i which atleast oe digit is repeated = = 80 Example # 23 How may fuctios ca be defied from a set A cotaiig 5 elemets to a set B havig 3 elemets. How may these are surjective fuctios. Solutio. Image of each elemet of A ca be take i 3 ways. Number of fuctios from A to B = 3 5 = 243. Number of ito fuctios from A to B = = 93. Number of oto fuctios = Fid the sum of all three digit umbers those ca be formed by usig the digits. 0, 1, 2, 3, 4. As How may fuctios ca be defied from a set A cotaiig 4 elemets to a set B cotaiig 5 elemets. How may of these are ijective fuctios. As. 625, I how may ways 5 persos ca eter ito a auditorium havig 4 etries. As Dearragemet : Number of ways i which '' letters ca be put i '' correspodig evelopes such that o letter goes to correct evelope is! ( 1) 1! 2! 3! 4!! Example # 24 I how may ways we ca put 5 writigs ito 5 correspodig evelopes so that o writig go to the correspodig evelope. Solutio. The problem is the umber of dearragemets of 5 digits This is equal to 5! = 44. 2! 3! 4! 5! Example # 25 Four slip of papers with the umbers 1, 2, 3, 4 writte o them are put i a box. They are draw oe by oe (without replacemet) at radom. I how may ways it ca happe that the ordial umber of atleast oe slip coicide with its ow umber. Solutio. Total umber of ways = 4! = 24. The umber of ways i which ordial umber of ay slip does ot coicide with its ow umber is the umber of dearragemets of 4 objects = 4! 2! 3! 4! = 9 Thus the required umber of ways. = 24 9 = 15 Self Practice Problems: 28. I a match colum questio, Colum cotai 10 questios ad Colum II cotai 10 aswers writte i some arbitrary order. I how may ways a studet ca aswer this questio so that exactly 6 of his matchigs are correct. As I how may ways we ca put 5 letters ito 5 correspodig evelopes so that atleast oe letter go to wrog evelope. As. 119 Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phoe : , page 5 of 20