PERMUTATION In probability we use many techniques for counting. ermutation and combination are one among them. In general, nermutation represents the arrangement of objects in different order and combination represents the way of selecting the objects ‘rom a group. ‘undamental Principle of Multiplication Ifa first event A can happen in 'm' ways, and a second event B can happen in 'n' ways, then the two event A and B in succession can happen in m x n ways. 16. Given 6 flags of different colour, how many eee can be generated if a signal requires 3 flags one below other. Oe192 BUSINESS Mar S MATHEM, Many S Solution 11" flag have 6 ways 2_}2™ flag have 5 ways 3 [93% flag has 4 ways + Number of different signals = 6.5 x4 =120 wayg Fundamental Principle of Addition If an event A can happen in 'm' ways and another ey happen in ‘n’ ways, then either of two events A or B can h (m + n) ways. ent B can, Happen in 17. There are 5 non veg counters and 3 veg counters in a meg In how many ways a peson can have lunch ? 7 Solution : Number of ways to have nonveg lunch = 5 Number of ways to have veg lunch = 3 Total number of ways to have either nonveg or veg lunch 3= 8 ways. 18 How many 4 digits numbers greater than 4000 can be formed by using the digits 1, 3, 5, 6, 0,7, 4. If (i) Digits are repeated (ii) digits are not repeated Solution: Given digits are 1, 3, 5, 6,0, 7,4 (i) When digits are repeated, since the number should be greater than 4000, the 1000" place can be filled in 4 ways (5, 6, 7 and 4) Number of ways of filling 100" place is 7 Number of ways of filling 10" place is 7 Number of ways of filling units place is 7 Total number of ways = 4x 7x 7x7 = 1,372 -1 = 1371 Here 1 is subtracted to remove 4,000 from the list a-7S, PERMUTATION, ors! 15, PERM COMBINATION anyp 1m 3 ‘ 5 are not repeated d iw he number > 4000 ce tht int? of filling, 1000" place =4 ie stilling, 100" place = 6 (7-1) so eo ing 1 place = 5 of filling, unit's place = 4 A ects 480 ot esl Notation any natural number, then factorial of nis the Product all is 1 pers less than or equal to n | num ol of n is denoted by{n or n! x weax (We) X (R= 2)X ve X21 ample: 4)= = 4x3x2x1=24 sample: ge (6=6x5x4x3x2x1=720 Note : @) nl=nxn-Axn-2x..x2x1 nl=n(n- 1)! nl= n(n - 1)(n - 2)! fi) = 111= 'S.Find the value of 7! Solution : 7!=7x6x5x4x3x2x1=5,040 Find the value of (y 10 (ii) 33} (iii) 51-3! (iv) 81+3! Solution: Wha Wxox darn gt. =210 MXAX3xZx1 af - PBN2IN 23x 2 =13,800 ——_—_ i BUSINESS, SS MATE, Ie (@x7VOxSNANINAXT + BK IQ) = 40,320 + 6 = 40,326 ntheg rain’ . 1,1 - y solution: Given 71* 8) o itmn * 1 4(1§) = 2 = 8 mo 2.1f nf = 20(n-2)! find n Solution: Given n! = 20 (n-2)! | n(n-1) (pa = 20 (02h n(n-l) = 20=5x4 = n= 5 Permutation Arrangement of objects of a set in definite order is called ‘permutation’. It defines the various ways of arranging the data or group of data in an order. In day to day life we face many dilematic situations 1" arrangement. In those cases permutation can be used to get the different ways of arrangement and on that analysis a optimum solution can be find out. Example: The three numbers 1, 2, 3 can be arranged in different ways ie., (1.2.3), (13,2), (21,3), (2,3,1}, (31,2) 13,21}PERMUTATION, COMBINATION AND 195 ce L2dean He rranged in 6 different ways if we consider apat atime. if we have toarrange any two numbers out of the three numbers, ahem the arrangement (permutation) will be, (12) (13) PA} (23) Ba Here n= 3, ie, total number of objects 2) = 6 ways. 1 = 2, ie., number of objects selected for permutation (arrangement) In general number of arrangement of 'n' objects taken ''at a time is "P, and the total permutation (ways of arrangements) can te calculated by using the formula, “P= Gay "P also denoted as P(n, r) o ina circular tray ? 32. Solution: Given n =7 Required number of permutation = (nq) _ =O 6x5X4X3N2x1= 729 ways, Bitls be seat Seate, re together, “Ung , seated around the 33.In how many ways 5 boys and § table, so that aumber of two girls a Solution: First 5 boys are table Number of boys = 5 B B x B Number of ways boys can be seated = (5.1) “po Fale 4y3x2x15 24 xh Now 5 girls can be arranged to sit in 5 places in 5 we X4x3x2x1=120 oo Total number of arrangement 24 x 120 = 2,880 ways \ COMBINATIONS Arrangement of objects are made or s considering the order or sequence is called as Tepresents the different ways of selection of selection does not matter. Example : If we have three select any elected with out "Combination’, i Of objects. Here order people A, B and C and we have to two, it represents combination. We can select AB, AC, or BC. These are the 3 possbilities of selection. Tfwe compare the above case with permutation the arrangement can be AB, AC, BA, BC, CA, CB in 6 ways. But in combination we have to select the objects without Considering the order. Here AB = BA, AC=CA and BC= CB Hence number of ways differs from permutation to combination Combination of n objects taken ' n ata time is denoted by "G or C(nr)EE - pxample wa IC, =5, ®C=12 Example (i) "Gt 10, example: © Cyo =! Ww) "Q* Example :* (vy) 8G =" Caer Example: "G, = "C, 9C, = °C; © "C="onarts (vi) "P.="Cxnt Example: ®p.="8Cx 7! 34, Evaluate °C, Solution: °C, g-Sys= 631* wed 35. Evaluate - 19 19) _ 19x 18x14 Coy a ST 36. 1f "C= "C,, Find the value of n. Solution: ze Solution: Given Wehave "C="C.sn=1+r ———Solution: Given "G.="C; =n=16+5=21 Ze 2 Mx 194 94 "y=" Cig= risiris! 3txael =Un2x19 = 7230 38. If *C,"C, = Solution: Given oe a 2n! 3/81 2n 2n-1) (2n-2) Qn-syf Besfsx2r nie) at ae = 44+4 =48 lo | 39.In how many ways a team of 11 players can be selected or formed out of 15 players ? 7 Solution: Given n= 15, r= 11 Number of teams q bya? ~ Fey = asian = Bil eps xp ne Oumnnenover emnt officers and 3 businessman. In how many ways ‘ommittee be formed ? cana olution: Out of 4 economists, two can be selected in *C., ways ut of 5 government offices, two can be selected in 5c, ways out of 3 business man, one can be selected in*C , ways otal number of selection according to above condition x 2G x5O7Q J 4x =| ehya| [234 BES» TxTx2xt *LFxePxT2xT = 6x10 x3 = 180 ways 7 = 7] 3 aya |X| 1 ‘4, Among 8 men and 6 women in a group, 4 has to be selected for an event such that in selection there should be 2 women. Ia how many ways the team can be selected. Solution : Since the group should have exactly 2 women. Out of 6 women 2 women can be selected in °C , ways Remaining 2 men out of 8 men can be selected in °C, ways. Total number of selection 8 V6 [ef leSre 42. A box contains 7 white balls and 5 black balls, (a) in how many ways 3 balls can be selected. (b) In how many ways 4 balls can be selected such that at Least 3 balls should be black. 28 x 15 = 420 ways 8c, x °C Solution : Given Total number of balls = 7white + Sblack =12 () When 3 balls are selected, r =3 ——HX 3x25} “2 way, (b) When 4 balls are selected, there should be atleas which has 2 possibilities, 3 blac ban, (i)3 black ball and 1 white ball or (i) All (4) are black balls and no White batt (1) 3 black balls out of 5 1 white ball out of 7= “C tc [She] =107~70 ways (ii) 4 black balls out of 5 ‘no white ball is selected = 7C 4 Bey x7C, X1=5 ways Total ways of selecting atleast 3 black basis = ether () or (i) =70 7 ways. 43. From seven Indian and four Germans a committee of five members is to be formed. In how many ways can this be done when the committee consists, (i) All Indians (ii) Atleast 2 Germans Solution : Total members = 7Indian + 4 German = 11 To form a committee of 5 members (i) If all members are Indians, n= 7, r =5 fi 7 |= Tx6x 4 xix gf =21 ways ———ee 2.Germa @? 4 ar of section = 4C umber n and 3 Indians 5 = 6x35 = 210 ways. 1 3 Germans and 2 Indians 0 number of section 4C,x = 4x21 = 84 ways, (o)4 Germans and 1 Indian ( number of section 4C x "C,=1x7=7 ways. Total number of ways of selecting atleast 2 Germans is 210 + 84+ 7 = 301 ways. us Verify whether "C, +"C,_, = "*1C. ifn =10 andr =4 Solution : Given n = 10 and r =4 6 WG = 7p lOt__ 2 10x9x8x7x6! _ 5 "CaM G™ (OAT! gta eT 7 20 © mig -10tIC, Me, = "GHG from (i) and (ii) mesh, = mC, 45.In a company 3 posts are vacant, and 20 candidates applied for this post. In how many ways can selection be made, ®c,,7 "= avg (}) If one particular candidate X is always selected. (ii) IFY is always excluded.ae a BUSINEsg, MA Ty 01 mber of px ATIcg Soluti pos number o APPlic Total number of ways of selection = 2 a= Bp 20-3)13) = 20x 198180171 <"TIN3x2x1 “L140 ways (i) If X is always selected (ie., X is fixed for one Post) Number of post remaining = 2 and numberof appli ‘ant = 19 Total number of ways of selection: 19! sory) = BN 18x17 19-392! T7! x2x7 “I71 ways (ii) If Y is always excluded Number of post emaining = 3 and numberof applicant = Total number of ways of selection = "C, 19! ! = sn - PERE 0 ays 46, In how many ways a party of 5 ladies and 4 gentlemen can be formed from 10 ladies and 8 men, if Mrs. A and Mrp refuses to join the same party ? Solution = Given number of ladies = 10, number of men = 8 5 ladies from 10 can be selected in = "°C, ways 4 gentlemen from 8 can be selected in = C,ways. Total number of ways of selection = "C5 x8Cy - a0 y5! x ean OX9KBX7X6Xx 5. BX7X6x5x4! N5x4x3x2x1 © dxdx3x 2x = 252.x 70 = 17,640 ways ero If Mrs. A and Mr. B joins the same party, then total number ways of selection of such party eee —gy OF SETS, PERMUTATION, COMBINATION AND 207 r if0 pp : { Remaining 4 ladies from 9 and 3 Bo Gee. || Gentlemen from 7 ot 7 = OAL (7-313! 6x5! 7 x4t - = BERIT GLIA = 126 «35 = 4410 ways Number of selection of party without Mrs. A and Mr. B together ( Total \ _ Number of party with Mrs. A and Mr.B) =\xjumber of selection} together i = 17,640 - 4,410 = 13,230 Note? 1, If there are 'n' number of people, then total number of different handshakes = "C, 2, If there are n non colinear points in a plane than, total number of triangles = "C,, Total number of straight line = G 47. Ina party there are 30 people. Each shakes hands with each other. What is the total number of handshakes ? Solution : Given n = 30 Total number of handshakes = "C, = A(n-1) _ 3029) _ =e Sy 435 48. What is the number of triangle that can be formed with 10 Points of a circle. Solution : Given: number of point n= 10 Total number of triangle = "C. = R(N-1) (n-2) _ 10(9) (8) _ ““3x2xT "6 = 2120 3 Rem ———4 Combinations 189 7 MODULE - 5 J Notations - Permutations of n different things - Ciruclar permutations - Permu tations a all different - Restricted permutations - Simple problems. Combinations My Problems based on formula pecans eeeceeee raderial notations: The product of the first n natural numbers namely 1, 2, 3, ....n is called factorial qorn factorial and is written as n! Thus on! = 1x2N3x x(n) x Ix2x3x4x5_ or Sx4x3x2x1 [iustration 5.1 | Show 10! Show that 1! 90 Solution: Y = 10x81 _ 99 —— illustration 5.2 — Solution: Heren= dre = f= SO 2 2 ways Circular Permutations These are related with arrangement of objects as in the case of a si fof members in table conference. Here the arrangeme, unless the order changes. itting arrange, nt does not Change un In circular permutations, the relative position of the other obj jects depend ot the objects placed first. It is only then the arran, . a) 1 \geMeNts of the gi objects is made other (n-1)! ways and not nt Thus the circular arrangement of 5 persons will be (n-1)! ie,, (5-1)! Tilustr: In how many ways 5 boys and 5 girls can be seated around a table so that no boys are adjacent. Solution: If the girl sits first the no. of ways girls can sit is (n-1)! ie, (Ge)! = 4! = 24 ways Now the boys can occupy 5 places 3 = 120 ways Total number off ways — 24x120 = 2880 Illustration 5.13 ind In how many ways 4 Kannadigas and 4 Maharastrians be seated at a r0U! table so that no two Kanndigas may be together. el ——Ss 4 Combinations 195 outatzons AO OS piution® met Kanai ways Kannadigas sit is as sit first no. of weit = (=D! = 6 ways ( f ways Maharastrians sit is he no. ren = 41 = 24 ways ate = 6x24 = 144 otal no. of Wa tations of things not all different: permul rhe number of permutations of n things of which P things are of one kind, q of a second kind, r things are of a third kind and the ° rest are differes tings ae ferent, then Find the number of permutations of the word ACCOUNTANT Solution: Here the word ACCOUNTANT has 10 letters Hence n = 10. It has 2A’s, 2C’s 2N’s, 27s. and the rest are different " 10! = 10x9x8; BRIAR = SOOT aan 2,26,800 {Illus Illustration 5.15 Find the number of permutations in the word ENGINEERING Solution: 0 The word ENGINEERING has 11 letters. -. n= 11 The word has 3E’s, 3N’s, 2G s, 2I's and the rest are different i! BTID! a111 0x9x8x7x6xBedx3x2x1 BxDx LxBK2x1XDXD 2K = 2,27,200 Find the number of arrangements that can be made with the ASSASSINATION, Worg Solution The word ASSASSINATION has 13 letters, with 3A’s, 48's, 2I's, 2N’s, and rest are different nt X= phq!ar! 13 Sdn! = 13x11 1x xxx 7xGxSxdxSx2n1 ‘BxDx txdxtxdx ixDxt xt = 1,08,10,800 Illustration 5.17 How many arrangements can be made with the letters of the word MATHEMATICS’ and in how many of them vowels occur together? Solution: (i) The word MATHEMATICS has 11 letters with 2M's, 2A’s, 21’s and rest are different Total number of arrangements nt X = piaghe! = TL TnlOdshxZxoxSuhyGuxd IHD! Bh = 49,89,600 (ii) There are 4 vowels in the word ‘MATHEMATICS’, namely AAIE. Two find the number of arrangements in which four vowels accur together, we need to consider the four vowels as one letter. Thus we are left with 8 letters of which 2M's, 27 and rest are different. ci ——4 Combinations n prox$xX2X1 = xx OT io = 10,080 tof 4 vowels, 2 are same ( out d Again ‘e A). Hence, total number of arrangements a! 2 12 | number of arrangements .10,080x12 = 1,20,969 Total jon 5.18 itustration 5.1 many ways can the letters of word ARRANGE be arranged? How In bet pese arrangements are there in which many of thes " ) The two R’S comes together? (i) The ) The two R's do not come together? ii (ii) The two R's and 2 A's come together? Solution: The word ARRANGE. has 7 letters with 24's, 2R’s and rest all different The number of permutations —5 1 = oko = XOsSadaSrPa = 1260 ways (\) The Two R’s come together —» 28's to be taken as ‘one. Hence the remaining letters are 6 out of which there are 24's, = 6! L 6x5x4x3; X= gy = Saou (iy = 360 TWo R's donot cone together —> Total arrangements - Arrangements where 2R’s come together *e, 1260 ~ 360 = 960 (iii) 2R’s and 2A’s come together Let us. take 2R’s and 2A’s as one letter. aoN 198 Quantitative r echni. ee ee Hence total letters — 3! Sxdxdu2xd i = 2OG2x1 = 129 x Restricted Permutations Most ofthe problems under permutations are with some restrictions. The res, may be several kinds like Tiction, a) One particular object is always included b) One particular object is never included «) Some objects are always together 4d) Some objects never occur together €) One particualr object occupied a particular place etc. Illustration 5.19 How many integers with different digits greater than 99999 are there with the restriction that 0 and 9 donot occur in the number? Solution: Since 0 and 9 are excluded, we have to form integers using 8 numbers ie, 1,2,3,4,5,6,7.8. As the number required is greater than 99,999, the required numbers should contain 6 digits, 7 digits and 8 digits. Number of 6 digit numbers = §P, = jis =8 = 5 = 20160 No. of 7 digits numbers No. of 8 digit nubers = = 40320 Total No, = 20160+40320+40320 =100800 justr Find the number of 4 digits numbers that can be formed using the digits 1.2345, if no digit is repeated? How many of these will be even? eel_ nd Combinations 19 a tation 0! pe ee tio =5 ol! 4 git numbers = No. e Styx 120 Numbers: Even numbers end with 2 or 4. Hence one unit dij 1 Even aining 4 digits should be arranged out of 3 digits ‘The rem 4p, = 24 ie, “Ps in No. of even numbers = 2x24 = 4g How many words, with or without mi jow many leaning can be made fr he word MONDAY, assuming that no | the w om the letters of letter is repeated, if {i 4 letters are used at a time, {i All letters are used at a time (ii) All letters are used, but first letter is a vowel Solution: There are 6 letters in ‘MONDAY’ () No. of words when 4 letters are used —5 ot ! Sp, = Gal = OSA _ 365 (No. of words when all letters are used —5 6! P= @a = ff = 720 Li) No. of words when all letters are used, but first letter is a vowel > First letter can be filled in 2 ways P,) used by using 5 letters 51 Sp, = 5 = 120 Total permantations = 2x20 = 240 These are 2 vowels (0,A), the remaining 5 places can be ged in a line In how Many different ways can 8 examination papers be arranged i % So that the best and worst Papers are never together? i __\Solution: Total no, of arrangements using all the papers = Be Sp, = (eay = 40320 Let vs consider fst and Worst POPES ORetNEr as 1 yn, 4 2 ers out of 7 can be arranged as 7 Remaining 6 paps c = 7 Coy = 9 TS together . Total arrangements where best and worst papers are n ! 0320 - 10800 = 30240 Ot together Total arrangements where best and worst paper = 81-(27!) Mlustration 5.23 Six papers ate set in an examination of which two are statist es. Ip, siferent ways canbe the papers be arranged so that the two statistics a SAFe otto? Solution: No. of arrangements of all 6 subjects = 6! If 2 statistics papers are to be kept together then the total arrangement Sis 51-2! ‘ (Since 2 statistics papers are taken as 1 unit and so remaining is 6-1 a The no. of arrangements in which 2 statistics papers are not together = Total arangements ~ Arrangements in which 2 statistics papers are togehy = 61 - Sh! = 720-240 = 480 Illustration 5.24 How many numbers of six digits can be formed from the digits 4,5,6,7,89 10 digit being repeated? How many of them are not divisible by 5? Solution: Total no. of six digits 6, = ay = 720 The digits divisible by 5 —» When 5 occurs in unit place. Hence the posiee YS is fixed in_unit place, The remaining 5 digits can be arranged —> esse oorn ; mm 4 Combinations 4 ee o ro 5p, = 5) ngements in which numbers are not divisible by 5 of arral are > rhe ° Total arrangements - Arrangements divisible by 5 = 720 - 120 = 600 nations: combina” ation, the objects are based on the order of the arr ‘angements where in permut’’ “der constitutes a different arrangement. But if order is not of any In in oF y zach chaNBe Tn it is_a problem of combination. Thus combinations are ee psequence? conse! pe made by taking some or all things at a time. In combin, can which ©! 7 . matte! joes not Joe pula for finding number of combinations —> Formula the groups ation, order n examination, 10 questions are set. In how In a many different ways can an examinee choose 7 questions? Solution: fn ca 10! We, = ohn = 10x9x8x7! SIXT = 10x9x8x7! “B2xIX7t = 120 Ulustration 5.26 How many committees of 5 members can be formed from 6 gentlemen and 4 ladies? Solution: _ 10! = (0-551 s = Wx%x8y7r6x5!_ © SxbGxdhst = 252 \ [Sm‘There are 12 persons including 4 ladies. tn how many ya be chosen 50 98 t0 include (} Ore lady (i) 4 Teast one gy * SOM ny ee Solution: {) Ifthe committee consists of 1 lady, then there should be 3 Total committees > acy af and | lag, 4 - wa x ym = AOS 43! = Shoal \ He = 5ox4 = 24 i) Committee consists of at least one lady. The combinations could be 3 4 ladies + 1 man and 3 ladies > 2 men amd 2 ladies > 3 men and 1 lady Acc, x 4c,)+'c, x 4C,) #8, x Ac) aint 1 (ee $| + [etiarets ” (ays) * Baye” 35! by |+[ gba 4! 4 BL nat /*Lax2 an |+[ sa =[ 4) q Ox a4) 4 | BxTE Ant], | Sx7aG! 4x2! | [Bx x65! 3! = UAT a JB oS] = 1+{8x4]+[28%6]+[56x4] = 143261684224 = 425 5.28 In how many ways can 5 books be selected out of 10 books, if 2 Mlustré books art (i) always selected (i) never selected eon —eC i me pinations tate peo soit always selected —> then 3 books should be selected from remaining got spooks are” aie aK )_ Bx7x6xS! 500 c= ANT Broaxt ~ 56 ver jenever selected > then all 5 books should be selected from remaining pit 2b00kS @ books 5! 4 t00 1 = BxTxGxS! = : 8c, * e5'3! = Beaaba! = 56 asset many ways 4 white and 3 black balls be selected from a box containing In how ivy 12 black balls? 16 white a” solution: wc, x C3 16 121 = Ged X C233 2 V6RISXI4XISAI! 12x11 0x9 FTE AXSx2XT OWXIx2xT = 1820 x 220 = 4,00,400 Illustration 5.30 From 6 boys and 4 girls, 5 are to be selected for a course. In how many way: this can be done if there must be exactly 2 girls? Solution: Koo 2 6 ! ~ © GZ = BSx43! SAAT A = 20x 6 120 ways t sts. In how ma 4 company, 4 posts fall vacant and 35 candidates apply for the posts. In hos Y ways selection can be made if a Pm Oy_ — ’~* Quantitative 7 : always included, (i) One particular candidate (ii) One particular candidate is always excluded, Solution: (i) One particular candidate is always included — Mo = Sl gy = GHD! S C= Ea Sima = 5984 (ii) One parc candidate is always excluded > = SagiGAgHs0! Mcy= pin SURE = 46376 Illustration 5.32 ‘A box contains 7 red, 6 white and 4 blue balls. How many selection of three balls can be made so that (i) none is red (ji) one of each colour? Solution: 10! (i) None is red + C, = (10-3131 = 1088 = 120 (ii) One each from each + Tex hc, 6 = 7h 4 = TA G1) TT =7x6x4 = 168 Mlustration 5.33 Ina group of 25 people, 15 are gents and 10 ladies. Find the number of ways of forming 15 people committee which should contain 9 gents. Solution: _ JS 10! = (15-9)/91 *(10-6)!6! 11x]099! , W0xBy7x6! x9! ADAG! 1051050inations 4 Combinat ions am 5 e value of ia wr, ©) C2) BC, “ws = gin = °AGS4 — 1680 w= Adis, = Wx9xBr7 xbxSrcdt = 151200 © Mc, = oda = 1BSSP = as w%,* wha = BH <5 Find the value of @7P, -*P, (@)°P,.51 (iy Wc, . oO Solution: 7 8! Py -8P3 = gy - hy TXOX5x4 ~ 8x7x6 840 - 336 = 504 Gi) 8Py 51 eayrax 5! BOSS 5x43. 001 70x120 8400 oe q (iii) Cg (worm x0! = 10x9x8x7! = Bane XI = 120x1 = 120 4 Paa | 06 Quantitative Teetnig S Migues Mlustration 5.36 How many permutations can be had out of 10 different objects by takin BA u —~ atime a {i) Hone object is to be taken every time (ii) 1 one object is not be taken every time Solution: (i) Ione object is to be taken at a time — P, = 0%) = Ox8x7 =504 (ii) One object is not be taken every time — 4 Ot)! oe) = 9x8x7x6 = 3024 linaseaneeieaol Hlustration 5.37 In how many ways the letter of the word FAILURE can be arranged if the 4 vowels are to remain together always. Solution: The word FAILURE’ has 4 vowels to be kept together. - to be taken as one letter Total letters in the word ‘FAILURE’ - 4 vowels and 3 consoments 4 vowels as one letter - Hence, 1+3 = 4 Hence, 4 letters are to be arranged themselves 4P, and 4 vowels are tobe arranged themselves 4P, 4p, x 4P, = 4! x 4! = 576 [Hlustra In how many ways 3 boys and 5 girls can be arranged so that 3 boys are always logether? Solution: These are 8 members = 3 boys and 5 girls. boss together as one - and the remaining 5 girls clpomeieieames as and Combinations tations 200 total 6 ence 6 u tal permutations °P,, Tota aD permutations of 3 boys themselves Be neta permutations °P, x 3P. = 6x3! 3 =720x6 = 4320 fiiustration 5.39 how many Ways can be letters of the word ‘COMMERCE’ be arranged? in ot Solution: . . ‘These are 2 C’s, 2M’s and 2R’s and Test all different in the word, COMMERCE, x ghaqixa! = asdhar = 5040 There are 11 questions in a paper of SSLC Examination. In how many ways an. examinee can select 6 questions? If quest ‘on No. 11 is made compulsory, in how many ways an examinee can select 6 questions? Solution: (@) The examinee has to select 6 questions out of 11, where there is no order Nee = qi, = UxlOwrsyx6 _ Co= (1-661 = ERRNO = 469 (HM questions no. 11 is made compulsory 0 Oe = Wxx8\7x6 _ (oss: = RRSP = 252 Mlustration 5.41 |. In how From 7 Englishmen and 4 Americans, a committee of 6 is tobe formed. In ho Many Ways can the committee be formed when (1) the committee contains exactly 2 Americans (ii) the committee contains at least 2 Americans ‘ PeQuantitatiy ee Tiling Solution: 2 4 WC x4, = Gaya eT Drorind! , dxgr2ai! = Rana? KP | = 35x3 = 210 (ii) At least 2 Americans > Possibilities > 2 Englishmen and 4 Americans 4 3 Englishmen and 3 Americans + 4 Englishmen and 2 Americans (Gx) +(c xc) +c, x‘c) Hy aay) *(rarar® aya) = (odpxradial | 2 oan) *( as rie) *(shar® 28) = (21x1)+(35x4)+(35x6) = 214140+210 = 371 EXERCISES 1. Find the number of permutations the letters of the word INDIA. (Ans.: 60) 2. Find the number of permutaions of the letters of the word EXAMINATION : (Ans. anBho) 3. How many words can be formed with the letters of the word TRIANGLE such that the word begins with N. (Ans.: 7!) 4. How many ways can 10 men stand in a row so that (2) Two particular men always together (>) Two particular men donot stay together |er revatio 10, u 12. 13. 4 15, and Combinations ng and Cor ny ways the letters of the word ‘daughter’ can be arrange, om gether? 'd keeping. Int the Ans. 3x6!) : any two digit numbers can be formed out How a is in each number (ii) 1 is not in e: that (i 3A si ach number a (Ans v, aut ° of the letters of the word ‘Cul pw that the permutations, ‘ulcutta’ is twice thy - utations of the letters of the word ‘America’ a permuta Find the value of (i) 8C, (i) C,. (iiy Ye) In how many ways a committee of 6 can be selected out of 9 (Ans: °¢y out of 13 wor PG xe) How many committees can be formed men and 4 men by taking 3 women and 2 men? (Ans If a student has to answer 8 uestions out of 10, then (i) in how many ways he can select 8 questions? (ii) If the first 3 questions are to be answer can select 8 questions? (Ans. ed, then in how many ways he Cy 7C5) (ti) One particular player must be excluded from each team? (Ans: 4C,,, Be) From 6 boys and 4 girls, 5 are to be selected for addmission for a selected Sourse. In how many ways can this be done if there must be exact 2 girls 5 6 (Ans.: °C,x4¢,) ; 7 How many words can be formed using all the letters of the word ‘BUSINESS’ conor