PERMUTATION AND COMBINATION ‘Suppose 4 members (say w; x); 2) went 10a cinema where they get only 3 tickets. So itis required w select 3 members out of 4 members (wx, 3: 2) we get the following selections waxy y=, yzw and zwr these 4 selections are called the ‘umber of combination of 4 members taken 3 at atime In symbols we wit this ase, ‘Now consider 4 letters: x, = Suppose we wish to arrange these leters taken 2 ata time, we get the following arrangements ex, wy: W325, J 2) J= 24 24 2y. These arrangements are called the numberof permutations of | “Fleters taken 2 at atime. In symbols we write thisas 4p, ‘Note that in arrangement order isis _yx, Where a in selection order isnot important Hence (wx) ‘same as yew, ywx and so itis egarded as only one selection. FACTORIAL OF A POSITIVE INTEGER The product of fist ‘n* natural numbers te, | andi represented bythe symbol | orn so [tet B [3123-6 [fe1a34.-24 Permutation and Combination 17 Also. =n(n=1).21 Ltn LOD min PERMUTATION ‘An arrangement of all or part ofa set of object permutation, Ifthe objects are arranged along a straight line, it i called a Hinear ‘permutation and ifthe objects are arranged around a circle then itis called 2 Circular permutation. ‘The number of permutations of ndistinct objects taken rata time ris My, =n I(m=2)u (FD This canbe proved by considering the number of ways of filling placesin [=2 andso on. some order is called row by any r objects ofthe m objects. Place 12 I r No.ofways: n (n=I) 2) rely “The first place canbe filed by any one ofthe °n” objects therefore in any cone of n ways. After the frst place the second place can be filled by any one of the remaining » ~ | objects and in (n-1) ways and so on. The * place can be filled by any one of the remaining n ~~ I) objects. So\in nr + | ways. By fundamental principle, the r places canbe filed by any of mobjects in My, = (01) (~ 2h (nr 1) ways, nr Ifn=2)ufn—r + hn 2d Now ty = (iaPXn—P— Don [onultipty and divide by (nr) (n= r1)..2.1] Kn "= Gent Note. The number of permutations of distint objects taken all ta time is ty, and by fundamental principle itis given by min—1)(n-2)..2.1 $0 My, = mn M(n—2).2.1 m= La ray 3.0 Ini elle factorial nLede [on [0 But from (1), =| WORKED EXAMPLES 1. Evaluate () 6, (i) 7p, (ii) 8p, 2 FM, = 720find n 70 Ls 123.an = 720. 1.2.3.4.5.6 = 720. n=6 Permutation and Combination us 3. fp, = 30 Find. n(n~1) = 30 r=n-30 = 0 0 ° “m6n + Sn—30 n(n ~6) +5(n~6) n= -Sorn=6, 6‘numbers can be formed with the digits 0, 1, 2,3, 4, 5? (repetitions not being allowed) How many ofthese are greater than 2000? Thousand Hundred Ten Unit TERAS eS Tae ‘The frst place (Thousand’s place) may be filled by any one ofthe given digits except, 030 this is done in $ ways. “The other3 places may be filled by the remaining 5 digits including zero in [5 Sx4x3x[2 Sn ways 155g awa. - Number of 4 digit numbers that can be formed with the digit 0, 1,2,3, 4, (repetitions not being allowed) = $ = 60 = 300 ways. Since we require the number greater than 2000, we can put 2, 3,4, 0r Sin thousand’s place so the thousand’s place canbe filled up in4 ways. The other 3 places can be filled in 5,, ways = 60 ways. 120 ‘Business Mavhematics [Number of 4 digit aumbers greater than 2000, that can be formed with the digits 0, 1,2, 3,4, = 4 « 60 = 240 ways 7. Imhow many different ways can S examination papers be arranged ina row, ‘0 thatthe best and the worst papers may never come together? Sol. ‘Without any reteton the S exam, papers canbe arranged among themselves in 5, =[5=5.4.3.2.1= 120 ways, Considering now the best and the worst papers as a single paper, we have only 3 + 1 =4 papers Now these 4 papers can be arranged taking all at atime in 4,, = |4 =4.3.2. 1-24 ways. Butin each of 24 ways, the best and the worst can be arranged among themselves in | 2 ways = 2 ways. ‘Total numberof arangemens where the best andthe worst papers always come together= [42 = 24 2= 48 ways, Hence the required number of arrangements where the best and the worst papers never come together = | S| 4] 2 = 120-48 = 72 ways. 8. There are 4 Kannada books, 3 Hindi books and 2 English books. In how ‘many ways can these be placed on a shelfif the books ofthe same language are tobe together? Sol. ‘Since the books on the same subject are to be together, let us consider the 4 ‘Kannada books as I unit,3 Hindi books as another unit and 2 English books as different unit. Then we have to arrange 3 different units. This car be done in 3p, ways = [3 =32%1=6 ways, But the 4 Kannada books remaining together can be arranged themselves in 4,, ways = [4 =4<3«2%1=24ways. ‘Next3 Hindi boos cn be aanged among themselves in 3, ways= [3 sways and 2 English books can be arranged among themselves in 2302COMBINATION ‘The number of ways of selection of» different things taken rata time (r W+6=n, sme, My Find Ma, Wehave m, =m. 3 rtin-nen My > 142m 30=n. ee eer) 4. 1f 20, = 20, Then find r 8 Business Mathematics 5. EM, = 105 find nSol. “a bee 3210 770 5 [10 2,2 ~T sata ‘ Pema ond Combination ns er 5 a aleass 7. ifm = 60and ”, = 10 then find mand r. “0 w-b = ie los = es Nowy = 6 (Given) or) = m= 1yin-2) = 60 Byispecion nes 8, A student has to answer 8 out of 10 questions in an examination, How ‘many choices has he, if he must answer the fist 3 question. Sol ‘There ate 10 questions of which a student must answer first 3 questions Remaining S questions +e has to answer 8 questions) can be selected among 10-37 questions in 7, ways. 1 Regd. numberofways = [Fas[3 7Tx6x|5 Bas 2 ways 9. Find the number of straight ines that ean be drawn fom 12 pointsof whieh 4 are collinear. Also find the numberof triangles that can be formed For these pointSol. ‘Two points are needed for a straight line. If none of the 3 given points are collinear then we would get '2,, straight lines. But 4 points are given to be collinear. So we would not get 4,, lines from these points instead we get only ‘one line containing all the 4 points. +. Number of straight lines = 12-4, +1 12 4 12-2[2 [4-2[2 12x IIx[ 10 43x] 2 [tox2 2 66-6+1 = 61 (8) We need 3 non collinear points for a triangle. If none of the 12 given points are collinear then we would get 12,, triangles. Since 4 points are given to be collinear, we would not get 4,, triangle from these points. +. No, oftriangles = 12, -4, le * [-3[3 [4-3/3 12x11x 10x] 9 4x3 = [932 3 = 220-4= 216. REMEMBER Alc. ree te LM ey = © Permutation of 1 objects of which p are of one kind, q are of another kind and so on is La Lely” Permutation and Combination . 131 Me, =A Me, =m, MH o ROR,