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1. ELEMENTARYPROBLEMS ON PERMUTATION & COMBINATION NOTE: USE FUNDAMENTAL PRINCIPLE OF COUNTING & ENJOY DOING THE FOLLOWING. Q.1 In howmanyways can clean& clouded (overcast) days occur in a week assuming that an entire dayis either cleanor clouded. [ Ans. 27 = 128 ] Q.2 Four visitorsA, B, C &D arrive at a town which has 5 hotels. In how manyways can they disperse themselves among5 hotels, if 4 hotels areusedto accommodate them. [ Ans. 5 . 4 . 3 . 2 = 120 ] Q.3 If the letters of the word “VARUN” are written in all possible ways and then are arranged as in a dictionary, then therank of the wordVARUN is : (A) 98 (B) 99 (C*) 100 (D) 101 Q.4 How manynatural numbers are theirfrom 1 to 1000 which have none oftheir digits repeated. [Hint: S D = 9 ; DD = 9 · 9 = 81 ; TD = 9 · 9 · 8 = 648] [Ans. 738] Q.5 3 different railwaypasses are allotted to 5 students. The number of ways this can be doneis : (A*) 60 (B) 20 (C) 15 (D) 10 Q.6 There are 6 roads between A & B and 4 roads between B & C. (i) In how manyways can one drive fromAto C bywayof B ? (ii) In how manyways can one drive fromAto C and back to A, passing through B on both trips ? (iii) Inhowmanyways can one drivethecirculartrip described in(ii) without using the same road more thanonce. [Ans. (i) 24 ; (ii) 576 ; (iii) 360 ] Q.7(i) How manycarnumber plates can bemade if each platecontains 2 different letters of english alphabet, followedby3different digits . [ Ans. 26 · 25 · 10 · 9 · 8 = 468000 ] (ii) Solve the problem, if the first digit cannot be 0. (Do not simplify) [Ans.26·25·9·9·8 = 421200] Q.8(i) Find the number of four letter word that can be formed from the letters of the word HISTORY. (each letter to be used at most once) [ Ans. 7 . 6 . 5 . 4 = 42 x 20 = 840 ] (ii) Howmanyof them containonlyconsonants ? [ Ans. 5 . 4 . 3 . 2 = 120 ] (iii) How manyof them begin & end in a consonant ? [ Ans. 5 . 4 . 4 . 4 = 400 ] (iv) How manyof them begin withavowel ? [Ans. 240 ] (v) Howmanycontainthe lettersY? [Ans. 480 ] (vi) How manybeginwith T & end ina vowel ? [Ans. 40 ] (vii) How manybeginwith T & alsocontain S ? [Ans. 60 ] (viii) How manycontainboth vowels ? [Ans. 240 ] Q.9 If repetitions are not permitted (i) How many3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 & 9 ? [Ans. 120 ] (ii) How manyof these are less than 400 ? [Ans. 40 ] (iii) How manyare even ? [Ans. 40 ] (iv) How many are odd ? [Ans. 80 ] (v) How manyaremultiples of5 ? [Ans. 20 ] CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-1
2. Q.10 Howmanytwodigitnumbersaretherein whichthetensdigitand theunitsdigit aredifferent and odd? [ Ans. 5 . 4 = 20 ] Q.11 Every telephone number consists of 7 digits. How many telephone numbers are there which do not include any other digits but 2 , 3, 5 & 7 ? [Ans. 47 ] Q.12(a) In how many ways can four passengers be accomodated in three railway carriages, if each carriage canaccomadate anynumber ofpassengers . (b) In how manyways four persons can be accomodated in 3 different chairs if each person can occupy onlyonechair [Ans. (a) 34 ; (b) 4.3.2 = 24 ] Q.13 How manyoddnumbers of five distinct significant digits can beformed with thedigits 0,1,2,3,4 ? [Hint:  3 × 3 × 2 × 1 × 2 = 36 Ans ] Q.14 How manyfourdigit numbers arethereall whosedigits are odd, if repetition ofdigits is allowed. [ Ans. = 54 ] Q.15 How manyfourdigit numbers aretherewhichare divisibleby2. [Ans. = 4500] Q.16 If all the letter of the word “MADHUR” are arranged by a lexsicographer then find the rank of the word “MADHUR” . [Ans. 362 ] Q.17 Find the number of 7 lettered palindromes which can be formed using the letters from the English alphabets. [Ans. 264] [Hint: A palindrome is a word or a phase that is the same whether you read is forward or backword. e.g. refer. ] Q.18 Number of ways in which 7 different colours in a rainbow can be arranged if green is always in the middle. [Ans 720 ] Q.19 Two cards are drawn one at a time & without replacement from a pack of 52 cards. Determine the numberof ways in which the two cards can be drawn ina definite order. [ Ans. 52 x 51 = 2652 ] Q.20 Findthe numberofwaysinwhichthelettersoftheword"MIRACLE"canbearrangedifvowelsalways occupythe odd places. Hint: [Ans:4·3·2·4·3·2·1=24·24=576 ] Q.21 Numbers of words which canbe formedusingall the letters oftheword"AKSHI", ifeach word begins withvowelorterminatesinvowel. [Ans: 2·24+2·24-2·6 = 84 ] Q.22 A letter lock consists of three rings each marked with 10 different letters. Find the numberof ways in which it is possible to makean unsuccessful attempts toopen the lock. [Ans: 103 - 1 = 999] Q.23 How many10 digit numbers can bemade with odd digits so that no two consecutivedigits are same. [Ans:5·49] Q.24 In how manyways can the letters of the word "CINEMA" be arranged so that the order of vowels do notchange. [Ans: 6 3 ! ! = 120] Q.25 How many natural numbers are there with the propertythat theycan be expressed as the sum of the cubes of two natural numbers in twodifferent ways . [Ans.Infinitelymany]
3. Q.1 Thenumberofarrangements whichcan bemadeusingall thelettersof theword LAUGHifthevowels are adjacent is: (A) 10 (B) 24 (C*) 48 (D) 120 [Hint: 4  2 !  3 ! = 48 ] Q.2 Thenumberofnatual numbers from 1000 to 9999(both inclusive)thatdo not haveall4 different digits is (A) 4048 (B*) 4464 (C) 4518 (D) 4536 [Hint: 7 · 8 · 9 · 9 10 9 3   ] Q.3 The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the conditionthat the digit 2 occurs exactlytwice in each number is (A*) 672 (B) 640 (C) 512 (D) none [Hint: Two blocks for filling 2can be selected in 7C2 ways and the digit 2 5 2 7 2 · C can be filled onlyin one wayother 5 blocks canbe filled in 25 ways.] Q.4 Out of seven consonants and four vowels, the number of words of six letters, formed by taking four consonants and two vowels is (Assume that each ordered group of letter is a word): (A) 210 (B) 462 (C*) 151200 (D) 332640 [Hint: 7C4 4C2 . 6 ! = 151200 ] Q.5 Allpossible threedigitsevennumberswhichcanbeformedwiththeconditionthatif5isoneofthedigit, then7is thenext digit is : (A) 5 (B) 325 (C) 345 (D*) 365 [Hint: 5 + 8.9.5 = 365 ] Q.6 For some natural N, the number of positive integral 'x' satisfyingthe equation , 1 ! + 2 ! + 3 ! + ...... + (x !) = (N)2 is : (A) none (B) one (C*) two (D) infinite [Hint: x = 1 & x = 3 ] Q.7 Thenumber ofsix digit numbers that can be formedfrom thedigits 1, 2,3, 4, 5, 6 & 7 so that digits do not repeat andthe terminal digits areeven is : (A) 144 (B) 72 (C) 288 (D*) 720 [Hint: 1 . . 3 . . 5 . . 7 3C2 · 2! · 5C4 · 4! = 6 x 120 = 720 ] CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-2
4. Q.8 In a certain strange language, words are written with letters from the following six-letter alphabet : A, G, K,N, R, U.Each word consists ofsix letters and none of the letters repeat. Each combination of thesesixletters is a wordinthis language.Theword"KANGUR"remains inthedictionaryat, (A*) 248th (B) 247th (C) 246th (D) 253rd [Sol. beginningwithAorG = 240 [18-12-2005, 13th] beginningwith 6 247th K A N G R U 248th K A N G U R ] Q.9 Considerthefivepointscomprisingofthevertices ofasquareandtheintersectionpointofits diagonals. Howmanytriangles canbeformedusing thesepoints? (A) 4 (B) 6 (C*) 8 (D) 10 [Hint: To form a triangle,3points out of 5 can be chosen in 5C3 =10 ways. But of these, the threepoints lying on the 2 diagonals will be collinear. So 10 – 2 = 8 triangles can be formed] Q.10 A5digitnumberdivisibleby3istobeformedusingthenumerals0,1,2,3,4&5withoutrepetition.The total number ofways this can bedone is : (A) 3125 (B) 600 (C) 240 (D*) 216 [Hint: reject 0 + reject 3  5! + 4 · 4! = 120 + 96 = 216 ] Q.11 Numberof9digits numbersdivisiblebynineusingthedigitsfrom0to9ifeachdigitisusedatmostonce is K . 8!, then K has the value equal to ______. [ Ans. 9 ! + 8 . 8 ! = 17 . 8 !  K = 17 ] [Hint: Case-I : reject '0' ; Case-II : reject 9 and take 0 , 1, 2, ....,8] Q.12 Numberofnaturalnumberslessthan1000anddivisibleby5canbeformedwiththetendigits,eachdigit not occuring morethan once in each number is ______ . [Hint : single digit = 1 ; two digit = 9 + 8 = 17 ; three digit = 72 + 64 = 136  Total = 154] Q.13 Threemenhave6different trousers, 5different shirts and4 differentcaps . Numberofdifferent ways in which theycan wear them is ______ . [ Ans. 6P3 . 5P3 . 4P3 ] Q.14 The number of 9 digit numbers that can be formed byusing the digits 1, 2, 3, 4 & 5 is : (A) 95 (B) 9! (C*) 59 (D) 9P5 Q.15 The number of arrangements of the letters 'abcd' in which neither a, b nor c, d come together is : (A) 6 (B) 12 (C) 16 (D*) none [Hint: 4 !  [ 3 ! 2 ! + 3 ! 2 !  2 ! 2 ! 2 ! ] = 8 ]
5. Q.1 Find the numberof ways in whichletters of the wordVALEDICTORYbe arranged so that the vowels maynever be separated. [Hint: VLDCTRY Y or 8! × 4! = 40320 × 24 = 967680 Ans ] [‘Valedictionmeans farewell aftergraduation from acollege.Valedictory:to takefarewell] Q.2 How manynumbers between 400 and 1000 (both exclusive)can be made with thedigits 2,3,4,5,6,0 if (a) repetitionofdigitsnot allowed. (b) repetitionofdigitsisallowed. [Hint: (a) 3×5×4 = 60 ; (b) 3×6×6 = 108 – 1 = 107 ] Q.3 The numberof ways inwhich5different books canbe distributed among10people if eachperson can get at most one book is : (A) 252 (B) 105 (C) 510 (D*) 10C5.5! [Hint: Select 5 boys in 10C5 and distribute 5 books in 5! ways hence 10C5. 5!] Q.4 Anewflagistobedesignedwithsixverticalstripsusingsomeorallofthecolouryellow,green,blueand red. Then,thenumberofways this canbe donesuch that no twoadjacent strips have thesame colouris (A*) 12 × 81 (B) 16 × 192 (C) 20 × 125 (D) 24 × 216 [Hint: 1st place can be filled in 4 ways 2nd place can be filled in 3 ways 3rd place can be filled in 3 ways and |||ly4th, 5th and 6th each can be filled in 3 ways. hence total ways = 4 × 35 = 12 × 81 ] Q.5 The9horizontal and 9 vertical lines on an 8 ×8 chessboard form'r'rectangles and 's'squares. The ratio r s in itslowest terms is (A) 6 1 (B*) 108 17 (C) 27 4 (D) none [Sol. no. of squares are [12 & 13th test (29-10-2005)] S = 12 + 22 + 32 + ........ + 82 = 6 ) 17 )( 9 ( 8 = 204 no. of rectangles r = 9C2 · 9C2 = 1296 hence r s = 1296 204 = 324 51 = 108 17 ] Q.6192/1 There are 720permutations of the digits 1, 2, 3, 4, 5,6. Suppose these permutations are arranged from smallest to largest numerical values, beginning from 1 2 3 4 5 6 and endingwith 6 5 4 3 2 1. (a) What numberfalls onthe 124th position? (b) What is the position of thenumber 321546? [Ans. (a) 213564, (b) 267th] [Sol. (a) digits 1, 2, 3, 4, 5, 6 no. of ways = 120 no. of number = 2 123rd finally 124th is = 213564 CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-3
6. (b) N = 321546 number ofnumbers beginningwith 1=120 number ofnumbers beginningwith 2=120 starting with 31 .............................. = 24 starting with 3214 .......................... = 2 finally = 1 hence N has 267th position ] Q.7 Astudent has toanswer10outof13questions inan examination. Thenumberofways inwhich he can answer ifhemust answeratleast 3ofthefirst fivequestions is : (A*) 276 (B) 267 (C) 80 (D) 1200 [Hint: 13C10  number of ways inwhich he can reject 3 questions from the first five or 13C10  5C3 = 286 – 10 = 276 or 5C3 . 8C7 + 5C4 . 8C6 + 5C5 . 8C5 = 276 ] [ Note that 5C3 . 10C7 is wrong ( cases repeat] Q.8 The number ofthreedigitnumbers havingonly twoconsecutivedigits identical is : (A) 153 (B*) 162 (C) 180 (D) 161 [Hint: when two consecutive digits are 11, 22, etc = 9 . 9 = 81 when two consecutive digits are 0 0 = 9 when two consecutive digits are 11, 22, 33, ... = 9 . 8 = 72  Total ] Q.9 Atelegraph hasxarms&eacharmiscapableof(x 1)distinctpositions,includingthepositionofrest. The total number of signals that can be made is ______ . [ Ans. (x  1)x  1 ] Q.10 Theinterioranglesofaregularpolygon measure150ºeach. Thenumberofdiagonals ofthepolygon is (A) 35 (B) 44 (C*) 54 (D) 78 [Hint: exterior angle = 30° Hence number of sides = 30 360 = 12  number of diagonals = 2 ) 3 12 ( 12  = 54 ] [Notethat sumofallexteriorangles ofapolygon=2and sumofalltheinterioranglesofapolygon=(2n-4)  2 ] Q.11 Numberofdifferentnaturalnumberswhicharesmallerthantwohundredmillion&usingonlythedigits 1 or 2 is : (A*) (3) . 28  2 (B) (3) . 28  1 (C) 2 (29  1) (D) none [Hint: Two hundred million = 2 x 108; (21+ 22 +23+24+25+26+27+28) + 28 = 766 ] Q.12 5 Indian & 5American couples meet at a party& shake hands . If no wife shakes hands with her own husband&noIndian wifeshakeshands with amale,then thenumberofhand shakes that takesplacein the partyis : (A) 95 (B) 110 (C*) 135 (D) 150 [Hint: 20C2  (50 + 5) = 135 ] Q.13 The number ofn digit numbers which consists of the digits 1 & 2 onlyifeach digit is to be used atleast once, is equal to 510 then n is equal to: (A) 7 (B) 8 (C*) 9 (D) 10 [Hint: (2 x 2 x .............2) n times-(when 1 or 2 is there at all the n places] [ Ans. 2n  2 ]
7. Q.14 Number ofsix digit numbers which have3 digits even &3 digits odd, ifeach digit is to be used atmost once is ______ . [Ans. 64800 ] [ when 0 is included (4C2 . 5C3 . 5 . 5 !) and when 0 is excluded (4C3 . 5C3 . 6 !) ] [Hint: alternatively, 5C3 · 5C3 · 6! since all digits 0, 1, 2, .........8, 9 are equally likely at all places  required number = 5 3 5 3 6 10 C C   ! · 9 digits or required number of ways = 5C3 . 5C3 . 6 ! – 4C2 . 5C3 . 5 . 5! ] Q.15 Find the number of 10 digit numbers using the digits 0, 1, 2, ....... 9 without repetition. How manyof these aredivisible by4. [Sol. Digit 0, 1, 2,........8, 9 Foranumbertobedivisibleby4the numberformedbylast twodigitsmustbe divisibleby4and can be 04, 08, 12, ........, 96 ; Total of such numbers = 24 Out of these 44 and 88 are to be rejected. (as repetition is not allowed) Hence accepted number of cases = 22 Out of these number of cases with '0' always include 04, 08, 20, 40, 60, 80 (six) no. ofsuch numbers with suchone of these as last two digits =6 · 8! ...(1) e.g. [ × × × × × × × × 04 ] no. of other numbers = 16 · 7 · 7! = 14 · 8! ....(2) e.g. [ × × × × × × × × 16 ]  Total number = 6 · 8! + 14 · 8! = (20) · 8! Ans. ]
8. Q.1 Thereare counters available in x different colours.The counters areall alikeexcept forthe colour.The total number ofarrangements consistingof ycounters, assuming sufficient number ofcounters of each colour, ifnoarrangement consists ofallcounters ofthesamecolouris : (A*) xy  x (B) xy  y (C) yx  x (D) yx  y Q.2 18 pointsare indicated ontheperimeter of atriangleABC (seefigure). Howmanytriangles are therewith verticesat thesepoints? (A) 331 (B) 408 (C) 710 (D*) 711 [Hint: 18C3 – 3· 7C3 = 816 – 105 = 711 ] [08-01-2005, 12th] Q.3 An English school and a Vernacular school are both under one superintendent. Suppose that the superintendentship, the fourteachership of English andVernacular school each, are vacant, if there be altogether11 candidatesfor theappointments, 3ofwhom applyexclusivelyforthesuperintendentship and2 exclusivelyfor theappointment inthe English school, the number of ways inwhich the different appointments can bedisposed of is : (A) 4320 (B) 268 (C) 1080 (D*) 25920 [Hint:  3C1 . 4C2 . 2 ! . 6 ! ] Q.4 A committee of5 is to bechosen from a group of 9 people. Number of ways in which it can be formed if two particular persons either serve together or not at all and two other particular persons refuse to servewitheach other, is (A*) 41 (B) 36 (C) 47 (D) 76 [Sol. 9 other 5 D , C AB ABincluded 7C3 – 5C1 = 30 (7C3 denotes any3 from (CD and 5 other) – no. of ways where ABexcluded 7C5 – 5C3 = 11 CD ais taken oneone from remaining) —— 41 ] [18-12-2005, 12th + 13th] Q.5 Aquestionpaperonmathematics consists oftwelvequestions divided intothreeparts A,Band C,each containingfour questions. Inhow manywayscan an examinee answer fivequestions, selectingatleast one from each part. (A*) 624 (B) 208 (C) 1248 (D) 2304 [Hint: 3 (4C2 . 4C2 . 4C1) + 3 (4C1 . 4C1 . 4C3) = 432 + 194 = 624 ] Alternative: [no of ways in which he does not select anyquestion from anyone section] 12C5 – 3 · 8C5 ] CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-4
9. Q.6 If m denotes the number of 5 digit numbers if each successive digits are in their descending order of magnitude and n is thecorresponding figure,when thedigits are in their ascending order ofmagnitude then (m – n) has the value (A) 10C4 (B*) 9C5 (C) 10C3 (D) 9C3 [Hint: m – n = 10C5 – 9C5 = 252 – 125 = 126 = 9C5 or 9C4] Q.7 There are m points on a straight lineAB & n points on the lineAC none of them being the pointA. Triangles areformed withthesepoints as vertices, when (i) Ais excluded (ii) Ais included.Theratio of number oftriangles in thetwocases is: (A*) m n m n    2 (B) m n m n     2 1 (C) 2 n m 2 n m     (D) m n m n ( ) ( ) ( )    1 1 1 [Hint: m C n C m C n C mn n m n m . . . . 2 2 2 2    ] Q.8 Numberofways inwhich9differentprizes begivento 5 students ifoneparticularboyreceives4 prizes and the rest of the students can get anynumbers of prizes, is : (A*) 9C4 . 210 (B) 9C5 . 54 (C) 4 . 45 (D) none [Hint: 4 prizes to be given to the particular boys can be selected in 9C4 ways . Remaining 5 prizes to the 4 students can be given 45 ways  total ways 9C4 . 45 = 9C4 . 210 ] Q.9 Inacertainalgebraical exercisebookthere are4examplesonarithmetical progressions, 5examples on permutation-combination and 6examples onbinomial theorem . Numberof ways ateacher can select for his pupils atleast one but not more than 2 examples from each of these sets, is ______. [Hint: ( 4C1 + 4C2) ( 5C1 + 5C2) ( 6C1 + 6C2) ] [Ans. 3150 ] [Alternatively: add one dummyexercies in each and compute 5C2 · 6C2 · 7C2 ] Q.10 The kindergartenteacherhas 25kids inherclass . Shetakes 5 ofthem at atime, tozoological garden as oftenas shecan,without takingthe same5 kids morethan once . Then thenumberofvisits, theteacher makes to the garden exceeds that of a kid by : (A) 25C5  24C5 (B*) 24C5 (C) 24C4 (D) none [Hint: Number of trips which exceeds  when one kid is never included  25C5  24C4 = 24C5 ] Q.11 n r n r n r r n C C C      1 0 1 is equal to : (A) n n n ( ) ( )   1 2 1 (B) n1 2 (C) n n ( ) 1 2 (D*) n 2 Q.12 Sevendifferentcoinsareto bedividedamongstthreepersons. Ifno twoofthepersonsreceivethesame numberofcoinsbuteachreceivesatleastonecoin &noneis leftover,thenthenumberofwaysinwhich thedivisionmaybemadeis (A) 420 (B*) 630 (C) 710 (D) none [Hint: 1, 2, 4 groups] [Ans. 7 1 2 4 ! ! ! ! × 3 ! ]
10. Q.13 Let therebe 9 fixed points on the circumference of a circle. Each ofthese points is joined to everyone of the remaining8 points bya straight line and the points are so positioned on the circumference that atmost 2 straight lines meet in anyinterior point of thecircle. The number ofsuch interior intersection pointsis: (A*) 126 (B) 351 (C) 756 (D) none of these [Hint: Any interior intersection point corresponds to 4 of the fixed points , namelythe 4 end points of the intersectingsegments. Conversely,any4labledpointsdetermine aquadrilateral, thediagonalsofwhich intersectonce withinthecircle.Thus theansweris A] Q.14 Thenumberof5digit numberssuchthat thesumoftheirdigits iseven is : (A) 50000 (B*) 45000 (C) 60000 (D) none [Hint: 2 10 9 4  ] Q.15 Aforecastis tobemadeoftheresultsoffivecricketmatches,eachofwhich canbewin, adrawora loss forIndianteam.Find (i) thenumberofdifferent possibleforecasts (ii) the number of forecasts containing 0, 1, 2, 3, 4 and 5 errors respectively [ Ans: 35 = 243 ; 1, 10, 40, 80, 80, 32 ] [Hint: N(0) = 1 ; N(1) = 2.5C4 ; N(2) 22. 5C3 ; N(3) 23. 5C2 ; N(4) = 24. 5C1 ; N(5) = 25 ]
11. Q.1 Thenumberofways in which 8 distinguishableapples canbedistributed among3 boyssuch that every boyshould get atleast 1 apple & atmost 4 apples is K · 7P3 where K has the value equal to : (A) 88 (B) 66 (C) 44 (D*) 22 [Hint: (4 3 1) ; (3 3 2) ; (4 2 2) ] Q.2 Awomenhas11closefriends.Findthenumberofwaysinwhichshecaninvite5ofthemtodinner,iftwo particular ofthem are not onspeaking terms & will not attend together. [Ans. 378] [Hint: 11C5  9C3 = 9C5 + 2 9C4 = 3 9C4 = 378 ] 5 to be selected      9C3 + 9C4 + 9C4 or 11C5 – 9C3 ] Q.3 A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it , so that therewill benocompletepair is : (A) 1920 (B) 200 (C) 110 (D*) 80 [Hint: 5C4 . 24 or 10 8 6 4 4 . . . ! = 80 ] Q.4 Numberofdifferentwaysinwhich8differentbookscan bedistributedamong3students,ifeachstudent receives atleast 2 books is ______. [Ans. 2940] [Hint: 8 books can be distributed in a group of (2, 2, 4) or (2, 3, 3). Number of groups are 8 2 2 4 2 8 2 3 3 2 ! ! ! ! ! ! ! ! ! !        and can be distributed in 3 ! ways ] Q.5203/1 Inhowmanydifferent ways agrandfatheralongwithtwo ofhisgrandsonsand fourgrand daughterscan be seated in a line for aphotograph so that heis always in the middle and the two grandsons are never adjacent to each other. [Ans. 528] [Sol. Total number of ways theycan sit = 6! × × × × × × no. of ways when the two grandsons are always adjacent = 4 · 2! · 4! = 192 where 4 denotes the no. of adjacent positions (2!) no. of ways in which two sons can be seated and 4!no. of ways inwhich the daughter canbe seated in the remaining places.  required no. of ways = 720 – 192 = 528 Ans ] Q.6 There are 10 seats in a doubledecker bus, 6 inthe lower deck and4 on the upper deck.Ten passengers board the bus, of them 3 refuse to go to the upper deck and 2 insist on going up. The number of ways in which the passengers can be accommodated is _____. (Assume all seats to be dulynumbered) [ Ans. 4C2. 2! 6C3. 3! 5! or 172800] Q.7 Findthenumberofpermutationsoftheword "AUROBIND" inwhichvowelsappearinanalphabetical order. [Ans. 8C4 · 4 !] [Hint: A,I,O,Utreatthemalike.Nowfindthearrangementof 8letters inwhich4alikeand4different= 8 4 ! ! ] Q.8 Thegreatest possiblenumberofpointsofintersectionof9different straightlines&9differentcirclesin aplaneis: (A) 117 (B) 153 (C*) 270 (D) none [Hint: 9C2 . 1 + 9C1 . 9C1 . 2 + 9C2 . 2 = 270 ] Q.9 Anoldmanwhiledialinga7digit telephonenumberremembers thatthefirst fourdigitsconsistsof one 1's, one 2's andtwo3's. Healsoremembers that thefifth digit is eithera4 or5 whilehas no memorising of the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number of distinct trials he has to tryto make sure that he dials the correct telephone number, is (A) 360 (B*) 240 (C) 216 (D) none CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-5
12. [Hint:       ! 2 ! 4       place fifth for ways 2       place 6 ways 10 th       place 7 way 1 th  th 7 3 3 2 1           x7 = 9 – x6 x6 can take 0 to 9 = 240 ] [12 & 13th test (09-10-2005)] Q.10 If as manymore words as possible be formed out of the letters of the word "DOGMATIC" then the numberofwords in which therelativeorderofvowels and consonants remain unchangedis ______. [Hint: I A 0 3! × 5! – 1 = 719 ] Q.11 Number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railway compartment iftwospecifiedpersons aretobealways included and occupyadjacent seatson thesame side, is 5! · (k) then k has the value equal to : (A) 2 (B) 4 (C*) 8 (D) none [Hint: includingthe twospecified people, 4 others canbeselectedin 5C4 ways. Thetwo adjacent seatscan be taken in 4 ways and the two specified people can be arranged in 2! ways, remaining 4 people can be arranged in 4! ways .  C4 . 4 . 2 ! 4 ! = 5 ! 8 = 8 . 5 !  C ] Q.12 Number of ways in which 9 different toys be distributed among 4 children belonging to different age groups in such a waythat distribution among the 3 elder children is even and the youngest one is to receive one toymore, is : (A)   5 8 2 ! (B) 9 2 ! (C*)   9 3 2 3 ! ! ! (D) none [Hint: distribution 2,2, 2 and 3 to the youngest . Now 3 toys for the youngest can be selected in 9C3 ways, remaining6 toys can bedivided into threeequal groups in   6 2 3 3 ! ! . ! way and can be distributed in 3 ! ways  9C3 .   6 2 3 ! ! =   9 3 2 3 ! ! ! ] Q.13 In an electionthree districts are tobe canvassed by 2, 3 &5 menrespectively. If10men volunteer,the number ofways theycan bealloted to the different districts is : (A*) 10 2 3 5 ! ! ! ! (B) 10 2 5 ! ! ! (C) 10 2 5 2 ! ( !) ! (D) 10 2 3 5 2 ! ( !) ! ! [Hint: number of groups of 2, 3, 5 = 10 2 3 5 ! ! ! ! & can be deputed only in one way ] Q.14 Let Pn denotes the number of ways in which three people can be selected out of 'n' people sitting in a row, if no two of them are consecutive. If , Pn + 1  Pn = 15 then the value of 'n' is : (A) 7 (B*) 8 (C) 9 (D) 10 [Hint: Pn = n  2C3 ; Pn + 1 = n  1C3 ; Hence n  1C3 - n  2C3 = 15 n  2C3 + n  2C2  n  2C3 = 15 or n  2C2 = 15  n = 8  C ] Q.15 The number ofpositive integers not greaterthan 100, which are not divisible by2, 3 or 5 is (A*) 26 (B) 18 (C) 31 (D) none [Hint: DrawaVenndiagram withcircles named A : the number of integers from 1 to 100 divisible by2 = 50 B : the number of integers from 0 to 100 divisible by3 = 33 C : the number of integers from 0 to 100 divisible by5 = 20 where theoverlapportions are thenumbers divisible byboth A & B , B & C, etc. (i.e. by 6, 10, 15 & 30). Thetotal numberofintegers not divisibleis then26. ] Use : n (A  B  C) = n (A) + n (B) + n(C) – n(A  B) – n (B  C) – n (C  A) + n (A  B  C) ]
13. Q.1 There are six periods in each working day of a school. Number of ways in which 5 subjects can be arranged if each subject is allotted at least one period and no period remains vacant is (A) 210 (B*) 1800 (C) 360 (D) 120 [Hint: 6C2 · 5C1 · 4! = 1800 S1 S2 S3 S4 S5 × × × × × note that at least one of the subject has to be repeated] Q.2 There are 10 red balls of different shades & 9 green balls of identical shades. Then the number of arrangingthem in a row so that no two green balls are togetheris (A) (10 !) . 11P9 (B*) (10 !) . 11C9 (C) 10 ! (D) 10 ! 9 ! Q.3 Numberofways inwhich n distinct objectscanbe keptintotwoidenticalboxesso thatno box remains empty, is ______. [Hint: Consider the boxes to be different for a moment. T1 can be kept in either of the boxes in 2 ways, silmilarlyfor all otherthings  Total ways = 2n but this includes when all the things are in B1 or B2  number of ways = 2n  2 Since the boxes are identical  2 2 2 n  = 2n  1  1 ] Q.4 Ashelfcontains20different books ofwhich4 areinsinglevolumeand theothersform sets of8, 5 and 3 volumes respectively. Number of ways in which the books may be arranged on the shelf, if the volumes of each set are together and in their due order is (A) ! 3 ! 5 ! 8 ! 20 (B) 7! (C*) 8! (D) 7 . 8! [Hint: Volume of eachset maybe indue order in two ways, either from left to right or from right to left. Now we have D1 D2 D3 D4 , = 7! × 2! . 2! . 2! = 8! ] Q.5 In a certain college at the B.Sc. examination, 3 candidates obtained first class honours in each of the followingsubjects:Physics,ChemistryandMaths,nocandidatesobtaininghonoursinmorethanonesubject; Numberofwaysinwhich9scholarshipsofdifferentvaluebeawardedtothe9candidatesifdueregardis to bepaid onlytotheplaces obtained bycandidates inanyone subjectis _____. [Ans: 1680 ] [Hint: 3 candidates in P; 3 in C and 3 in M. Now 9 scholarships can be divided into three groups in ! 3 . ) ! 3 ( ! 9 3 ways and distributed to P, C, M in 3 ) ! 3 ( ! 9 ways ] Q.6 Thereare12differentmarblestobedividedbetweentwo childrenintheratio1:2.Thenumberofways it can be done is : (A*) 990 (B) 495 (C) 600 (D) none [Hint: 12 4 8 ! ! !  2 ! = Ans. : 990 ] Q.7 All the five digits number in which each successive digit exceeds its predecessor are arranged in the increasingorder oftheirmagnitude. The 97th numberinthelist does notcontain thedigit (A) 4 (B*) 5 (C) 7 (D) 8 CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-6
14. [Sol. All the possible number are 9C5 (none containing the digit 0)= 126 Total numbers starting with 1 = 8C4 = 70 (using 2, 3, 4, 5, 6, 7, 8, 9) Total starting with 23 = 6C3 = 20 (4, 5, 6, 7, 8, 9) Total starting with 245 = 4C2 = 6 (6, 7, 8, 9) 97th number = ] Q.8 The number ofcombination of 16 things, 8 of which are alike and the rest different, taken 8 at a time is ______. [Ans. 256] [Hint: AAAAAAAA D1 D2 ......... D8  8C0 + 8C1 + 8C2 + ...... + 8C8 = 256 ] Q.9 Thenumberofdifferentwaysinwhichfive'dashes'andeight'dots'can bearranged, usingonlysevenof these 13 'dashes' & 'dots' is : (A) 1287 (B) 119 (C*) 120 (D) 1235520 [Hint: 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 ] Q.10 A gentleman invites a party of m + n (m  n) friends to a dinner & places m at one table T1 and n at another table T2 , the table being round . If not all people shall have the same neighbour in any two arrangement, then thenumber of ways in whichhe can arrange theguests, is (A*) ( ) ! m n mn  4 (B) 1 2 ( ) ! m n mn  (C) 2 ( ) ! m n mn  (D) none [Hint:     2 )! 1 n ( . 2 )! 1 m ( ! n ! m )! n m ( (A) ] Q.11 There are nidentical red balls &m identical green balls . The number ofdifferent linear arrangements consistingof"nredballs but not necessarilyall thegreen balls"isxCy then (A) x = m + n , y = m (B*) x = m + n + 1 , y = m (C) x = m + n + 1 , y = m + 1 (D) x = m + n , y = n [Hint: Put one more red ball & find the arrangement of n + 1 red and m green balls = m + n + 1Cm ] Q.12 In how manyways the letters of the word “COMBINATORICS” can be arranged if (i) All thevowels are always groupedtogether to form acontiguous block. (ii) All vowels andall consonants arealphabeticallyordered. [Ans. (i) 3 ) ! 2 ( ) ! 5 )( ! 9 ( ; (ii) ) ! 5 )( ! 8 ( ) ! 13 ( ] Q.13 Number of different words that can be formed using all the letters of the word "DEEPMALA" if two vowels are together and the othertwo are also together but separated from thefirst two is (A) 960 (B) 1200 (C) 2160 (D*) 1440 [Hint: | D | P |M | L| can be arranged in 4! ways & the two gaps out of 5 gaps can be selected in 5C2 ways. AA or EE or A E or AE can be placed in ! 2 ! 2 ! 4 = 6 ways.  4 ! · 5C2 · ! 2 ! 2 ! 4 = 1440 ]
16. Q.19 How manyfive digit numbers can be formed using 2, 3, 8, 7, 5 exactlyonce such that the number is divisibleby125? (A) 0 (B) 1 (C*) 4 (D) 3 [Hint: Ifthenumberformedbylastthreedigitsofanumberisdivisibleby125 thenthenumberitselfisdivisible by125. Hence just find thethree digits numbers which are divisible by125 and are formed byusing2, 3, 8, 7, 5 exactlyonce. 375 and 875 are the onlytwo numbers to satisfying both these conditions. The possible numbers are 28375, 82375, 23875 and 32875. Hence answer is (C). ] Q.20 Six people are going to sit in a row on a bench.Aand B are adjacent. C does not want to sit adjacent to D. E and F can sit anywhere. Number of ways in which these six people can be seated, is (A) 200 (B*) 144 (C) 120 (D) 56 [Hint: ; C and D separated ; E and F any where and E, F can be seated in 3! 2! no. of gaps are 4 | |E |F | C D can be seated in 4C2 · 2! Total ways 3! · 2! · 4C2 · 2! = 144 Ans. ]
17. Q.1 Number of ways in which n things of which r alike & the rest different can be arranged in a circle distinguishingbetweenclockwiseandanticlockwisearrangement,is: (A) ( ) ! ! n r r   1 (B) ( ) ! n r  1 (C) ( ) ! ( ) ! n r   1 1 (D*) ( ) ! ! n r  1 [Hint: x . r ! = (n  1) !  x =   n r  1 ! ! ] Q.2 Thenumberofwaysofarranging2mwhite&2nredcountersinstraightlinesothat eacharrangementis symmetrical withrespect to a central mark is ______. (assume that all counters are alike except for the colour) [Ans. ( ) ! ! ! m n n m  ] Q.3 Numberofwaysinwhichfourdifferenttoysandfiveindistinguishablemarblescanbedistributedbetween Amar,AkbarandAnthony, ifeach child receives atleast one toyand one marble, is (A) 42 (B) 100 (C) 150 (D*) 216 [Sol. Toys in group 1 1 2  ! 3 ! 2 ! 2 ! 1 ! 1 ! 4  = 36 Marbles O O O O   = 4C2 = 6  Total ways = 36 × 6 = 216 ] Q.4 Delegates from 9 countries includes countriesA, B, C, D are to be seated in a row . The number of possibleseatingarrangements,when thedelegates ofthecountriesAand Bareto beseatednext to each other and the delegates of the countries C and D are not to be seated next to each other is : (A) 10080 (B) 5040 (C) 3360 (D*) 60480 [Hint: | AB | E | F | G | H | I  7C2 2 ! 6 ! 2 ! = 60480 ] Q.5 There are 12 guests at a dinner party. Supposingthat the master and mistress of the house have fixed seats opposite oneanother, and that thereare two specified guests who must always, beplaced next to one another; the number ofways in which the companycan be placed, is: (A*) 20 . 10 ! (B) 22 . 10 ! (C) 44 . 10 ! (D) none [Hint: 6 places on either sides  G1G2 will have 5 places each on either side and can be seated in 2 ways  10 × 2! × 10! Ans ] Q.6 Let Pn denotes the number of ways of selecting 3 people out of 'n' sitting in a row, if no two of them are consecutive and Qn is the corresponding figure when theyare in a circle. If Pn  Qn =6, then 'n' is equal to : (A) 8 (B) 9 (C*) 10 (D) 12 [Hint: Pn = n  2C3 ; Qn = nC3  [ n + n (n  4) ] or Qn = n n C C 1 4 2 3 .  Pn  Qn = 6  n = 10 ] CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-7
18. Q.7207/1 Define a 'good word' as a sequence of letters that consists onlyof the lettersA, B and C and in which AneverimmidiatelyfollowedbyB, Bis neverimmediatelyfollowedbyC, and Cis neverimmediately followed byA.If the number of n-letter good words are 384, find the value of n. [Ans. n = 8 ] [Sol. There are 3 choices for the first letter and two choices for each subsequent letters. Henceusingfundamentalprinciple number of good words = 3 · 2n–1 = 384 2n–1 = 128 = 27 n = 8 Ans. ] Q.8 Six marriedcoupleare sittingina room. Findthenumber of ways in which 4 people can be selected so that (a) theydo not form a couple (b) theyformexactlyonecouple (c) theyform at least one couple (d) theyformatmost one couple [Ans. 240, 240, 255, 480] [Hint: 12C4 = 6C2 + 6C1 · 5C2 · 24 + 6C4 · 24 ] Q.9 In a conference 10 speakers are present . If S1 wants to speak before S2 & S2 wants to speak after S3, then the number of ways all the 10 speakers can give their speeches with the above restriction if the remainingsevenspeakers have no objection to speak at anynumber is (A) 10C3 (B) 10P8 (C) 10P3 (D*) 10 3 ! [Hint: S1 S2 S3 or S3 S1 S2  10C3 . 2 . 7 ! ] Q.10 On a Railwayroute from Kota to Bina there are 12 stations.A booking clerk is to be deputed for each of these stations out of 12 candidates of whom five are Marathis, four are Oriyas and the rest are Begalis.Thenumberofways ofdeputingthepersonson thesestationsso thatno twoBengali’s serveon two consecutive stations, is _____________. ( Persons of the same religieon are not to be distinguished) [Hint: B B B X | X | X | X | X | X | X | X | X | Select 3 gaps = 10C3 = 120  number of arrangements = 120 × ! 4 . ! 5 ! 9 = 120 × 126 = 15120Ans] Q.11 Let m denote the number of ways in which 4 different books are distributed among 10 persons, each receivingnoneorone onlyandlet n denote thenumberofways ofdistribution if thebooks areall alike. Then: (A) m = 4n (B) n = 4m (C*) m = 24n (D) none [Hint: m = 10C4 . 4 ! and n = 10C4 ] Q.12 The numberofways inwhichwecan arrangen ladies & n gentlemen ataround table so that2 ladies or 2 gentlemen maynot sit next to oneanother is : (A) (n  1) ! (n  2)! (B*) (n)! (n1)! (C) (n +1) ! (n)! (D) none [Hint: arrangethem alternatelyon thecircle] Q.13 Thenumberofways in which10identical apples canbe distributedamong6 children sothat eachchild receives atleast one apple is : (A*) 126 (B) 252 (C) 378 (D) none of these Q.14 Thenumberofall possibleselectionsofoneormorequestions from 10givenquestions,eachequestion havinganalternativeis: (A) 310 (B) 210  1 (C*) 310  1 (D) 210 [Hint: 1st question can be selected in three ways and so on ]
19. Q.15 The number of ways in which 14men be partitioned into 6 committees where two of the committees contain 3 men & the others contain 2 men eachis : (A) 14 3 2 2 4 ! ( !) ( !) (B) 14 3 2 2 5 ! ( !) ( !) (C) 14 4 3 2 2 4 ! ! ( !) . ( !) (D*) 14 2 3 4 5 2 ! ( !) . ( !) . ! Q.16 The number of divisors of the number 21600 is _____ and the sum of these divisors is ____. [ Ans. 72, 78120 ] Q.17 10 IIT & 2 PET students sit in a row . The number of ways in which exactly3 IIT students sit between 2 PET student is ______ . [ Ans. 8 . 2 ! 10 ! = 16 . 10 ! or 10C3 . 3 ! 2 ! . 8 ! ] [Hint: 10 IIT students T1, T2, ..... T10 can be arranged in 10 ! ways . Now the number of ways in which two PET studentcanbeplaced will beequal to thenumber ofways inwhich 3 consecutiveIITstudents can be taken i.e. in 8 ways and can be arranged in two ways  (10!) (8!) (2!) . Alternatively 3 IIT student can be selected in 10C3 ways. Now each selection of 3 IIT and 2 PET students in P1 T1 T2 T3 P2 can be arranged in (2!) (3!) ways. Call this box X. Now this X and the remaining& IITstudents can be arrangedin 8! ways  Total ways 10C3 ( !) (3 !) (8 !) ] Q.18 The number ofways of choosing a committee of 2 women& 3 men from 5 women & 6 men, if Mr.A refuses toserveonthe committeeif Mr. Bisamember&Mr.B can onlyserve,ifMiss C isthemember ofthecommittee,is (A) 60 (B) 84 (C*) 124 (D) none [Hint: ; ; (i) Miss C is taken (a) B included  Aexcluded  4C1 . 4C2 = 24 (b) B excluded  4C1 . 5C3 = 40 (ii) Miss C is not taken  B does not comes ; 4C2 . 5C3 = 60  Total = 124 Alt. Total  [A, B, C present +A,B present & C absent + B present &A, C absent] Alternatively : Case 1 : Mr. 'B' is present  'A ' is excluded & ' C ' included Hence number of ways = 4C2 . 4C1 = 24 Case 2 : Mr. 'B ' is absent  no constraint Hence number of ways = 5C3 . 5C2 = 100 Total = 124 ] Q.19 Six personsA, B, C, D, E and F are to be seated at a circular table. The number of ways this can be done ifAmust have eitherB or C on his right and Bmust have either C or D on his right is : (A) 36 (B) 12 (C) 24 (D*) 18 [Hint: whenAhas B or C to his right we haveAB orAC when B has C or D to his right we have BC or BD Thus  we must haveABC orABD orAC and BD for ABC D, E, F on a circle number of ways = 3 ! = 6 for ABD C, E, F on a circle number of ways = 3 ! = 6 for AC , BD E, F the number of ways = 3 ! = 6  Total = 18 ] Q.20 Thereare2identical whiteballs, 3identicalred balls and 4green balls of different shades. Thenumber of ways inwhich theycan bearranged in arowso that atleast one ball is separated from theballs of the samecolour,is : (A*) 6 (7 !  4 !) (B) 7 (6 !  4 !) (C) 8 !  5 ! (D) none [Hint: 9 2 3 ! ! ! – number of ways when balls of the same colour are together = 9 2 3 ! ! !  3! 4! = 6(7!  4!) ]
20. Q.21 Boxes numbered 1, 2, 3, 4and 5 are kept in a row, andtheyare necessarilyto be filled with eithera red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then how manydifferent arrangements arepossible, giventhat theballs ofa given colour areexactlyidenticalin all respects? (A) 8 (B) 10 (C*) 13 (D) 22 [Hint: Justifywithatreediagram or alternatively: 0 B  R R R R R (1) = 1 1 B  | R | R | R | R | ( 5C1 ) = 5 2 B  | R | R | R | ( 4C2 ) = 6 3 B  | R | R | ( 3C3 ) = 1 —————————— Total = 13 ] Q.22 In howmanyways can theeight directors, theVice-chairmanand the Chairman ofa firm be seatedat a round-table , ifthe chairman has to sit between theVice- chairman and a director? (A) 9 ! × 2 (B*) 2 × 8 ! (C) 2 × 7 ! (D) 8 ! [Hint: LettheVice-chairmanandtheChairmanformsform1unit.Alongwith theeightdirectors,wenowhave toarrange9different units inacircle.This canbedonein 8!ways.At thesametime, theVice-chairman and the Chairman can be arrangedin twodifferent ways. Thereforethe total numberof ways is 2 × 8! ] Q.23 Howmanyfivedigitnumbers canbeformedfrom 1,2,3, 4, 5(without repetition), when thedigit atthe unit place must be greater than that in the tenth place? (A) 54 (B*) 60 (C) 5!/ 3 (D) 2 × 4! [Hint: Thenumbers shouldbeformedfrom 1,2, 3,4,5(withoutrepetition),such that thedigitattheunitplace must be greaterthan that in thetenth place.Tenth placehas five options. If 5 is at the tenth place the digit at the unit’s place cannot be filled bythe digit greater than that at the tenthplace. If 4 is at thetenth place, then theunit’s place has onlyoption of 5, whileotherthreeplaces canbefilled up in 3! ways. If 3is at thetenthplace, thentheunitsplacecanbefilledupby4 or5i.e. in two ways.Whileotherthree places canbe filled up in 3! ways. If 2 is at the tenth place, then the unit’s place can be filled up by3, 4 or 5, i.e. in 3 ways. While other three places can be filled up in 3! ways. If 1 is at the tenth place, then anypther four places can be filled up in 4! ways. Thus the total number of numbers satisfying the given conditions is 0 + 3! + 2(3!) + 3(3!) + 4! = 60. Hence the answer is (B) ] Q.24 Fifty college teachers are surveyed as to their possession of colour TV, VCR and tape recorder. Of them, 22 owncolourTV, 15 ownVCR and 14 own taperecorders. Nine of thesecollege teachers own exactlytwo items out of colourTV,VCR and tape recorders ; and, one college teacher owns all three. how manyof the 50 college teachers own none of three, colourTV,VCR or tape recorder? (A) 4 (B) 9 (C*) 10 (D) 11 [Hint: BydrawingaVenn diagram , we see that thenumber of teachers with some possession = [ 22 + 15 + 14 – 9 – 2(1)] = 40 .  the number of people havig no possesion = 50 – 40 = 10 ] Q.25 The number of ways in which a mixed double tennis game can be arranged from amongst 9 married couple ifno husband &wifeplays in thesamegameis : (A) 756 (B) 3024 (C*) 1512 (D) 6048 [Hint: 9C2 . 7C2 . 2 ! = 1512 ]
21. Choose the correct alternative (only one is correct): Q.1 The combinatorial coefficientC(n, r) can not beequal to the (A) number ofpossible subsets of rmembers from a set of n distinct members. (B)numberofpossible binarymessages of length n withexactlyr 1's. (C*) number of non decreasing 2-D paths from the latticepoint (0, 0) to (r, n). (D) number of ways of selecting r things out of n different things when a particular thing is always included plus the number of ways ofselecting 'r' things out of n, when a particular thing is always excluded. [Hint: In C it should be (r, n + r) ; (D) nCr = n – 1Cr – 1 + n – 1Cr ] Q.2 Given11points,ofwhich5lieononecircle,otherthanthese5,no4lieononecircle.Thenthemaximum numberof circles that can be drawnso that each contains atleast three of the given points is : (A) 216 (B*) 156 (C) 172 (D) none [Hint:  5C2 . 6C1 + 6C2 5C1 + 6C3 + 1 = 156 alternatively 11C3  5C3 + 1] Q.3 Messages areconveyedbyarranging4white, 1blueand 3redflags on apole. Flags ofthesamecolour arealike.Ifamessage istransmittedbytheorderin which thecoloursarearrangedthenthetotal number of messages that can be transmitted ifexactly6 flags areused is : (A) 45 (B) 65 (C) 125 (D*) 185 [Hint: consider, 4 alike + 2 others alike , 4A + 2 different 3 A + 3 OA and 3 A + 2 OA + 1 different  15 + 30 + 20 + 120 ] Q.4 The number of ways of arranging the lettersAAAAA, BBB, CCC, D, EE & F in a row if the letters C are separated from one another is : (A) 13 5 3 3 2 ! ! ! ! ! (B) 14 3 3 2 ! ! ! ! (C*) 11 . 13 6 ! ! (D) none Q.5 Apersonwrites letters tohis 5friends and addressesthecorrespondingenvelopes. Number of ways in whichtheletterscanbeplacedinthe envelope,sothatatleasttwoof themareinthewrongenvelopes,is, (A) 1 (B) 2 (C) 118 (D*) 119 [Hint: 5! = W 4 and R 1 + W 3 and R 2 + W 2 and R 3 +      way one used none R 5 4 + all 5 wrong hence at least 2 in wrong = 120 – 1 = 119 ] Q.6 You are given an unlimited supplyof each of the digits 1, 2, 3 or 4. Using only these four digits, you constructn digitnumbers. Such n digitnumbers will be calledLE GITIMATE if itcontains the digit 1 either aneven number times ornot at all. Numberof n digitlegitimate numbers are (A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n – 1(2n + 1) [Sol. × × ×............× (n places) Totalnumbers = 3n + nC2 · 3n – 2 + nC4 · 3n – 4 + .......... (nC2 indicates selection of places) = 2 ) 1 3 ( ) 1 3 ( n n    = 2 1 [4n + 2n] = 22n – 1 + 2n – 1 = 2n – 1(2n + 1) ] CLASS : XIII (XYZ) Special DPP on Permutation and Combination DPP. NO.-8
22. Q.7 Two classroomsAand B havingcapacityof 25 and(n–25) seats respectively.An denotes thenumber of possible seatingarrangements of room 'A', when 'n' students are to be seated in these rooms, starting from room 'A'which is tobe filled up full to its capacity. If An –An–1 = 25! (49C25) then 'n' equals (A*) 50 (B) 48 (C) 49 (D) 51 [Hint: Given An = nC25 · 25! ; An – 1 = n–1C25 · 25! hence nC25. 25! – n–1C25 . 25! = 25! 49C25 or n–1C25 + n–1C24 – n–1C25 = 49C24  n – 1 = 49  n = 50  n–1C24 = 49C24  n = 50] Q.8 LetA,B,C,D,Eand Fbethevertices ofaregularhexagon with Gasits centre. IfSdenotes thenumber ofstraightlinesandTdenotesthenumberoftriangleswhichcanbehadbyjoiningthese7pointsthenthe ordered pair (S, T), is (A) (18, 32) (B) (12, 26) (C*) (15, 32) (D) (15, 26) Q.9 The number oftimes thedigit 3will be writtenwhen listingtheintegersfrom 1 to 1000is : (A*) 300 (B) 269 (C) 271 (D) 302 [Hint: A three digit block from 000 to 999 means 1000 numbers, each number constituting 3digits.Hence,total digits whichwe have to writeis 3000.Since thetotal numberof digitsused are 10 (0to9) andnodigitis filledpreferentialy, hence the number of times we write 3 = 3000 10 = 30. Alternatively :anyoneblock can be selectedin 3C1 ways and the digit 3 can befilled in it.Now the remaining two blocks can be filled in 9  9 = 81 ways (excludingthe digit 3) Total ways it can be done = 3C1 . 9 . 9 = 243 Similarly any two blocks can be selected in 3C2 ways and the digit 3 can be filled in both of them. Remainingoneblock can be filledin9 ways . Total ways this can be done = 3C2 . 2 . 9 When all the blocks are taken we have 3C3 . 3 Thus the total = (3C1 . 92) 1 + (3C2 . 9) 2 + 3C3 . 3 = 300 ] Q.10 Distinct 3digit numbers are formed usingonlythe digits 1, 2, 3and 4with each digit used at most once ineach numberthus formed. Thesum ofall possiblenumbersso formed is (A*) 6660 (B) 3330 (C) 2220 (D) none [Hint: all possible= 24 6(1 + 2 + 3 + 4)(1 + 10 + 102) = 6 · 10 · 111 = 6660 reject 1 or 2 or 3 or 4 ] Q.11 There are counters available in 3 different colours (atleast four of each colour). Counters are all alike except for the colour. If 'm' denotes the number of arrangements of four counters if no arrangement consists of counters ofsame colour and 'n' denotesthe corresponding figure when everyarrangement consists ofcounters of eachcolour, then : (A) m = 2 n (B*) 6 m = 13 n (C) 3 m = 5 n (D) 5 m = 3 n [Hint: m = 34  3 = 78 n = 34      3 2 2 3 4   = 81  45 = 36 Hence m n = 78 36 = 13 6  6 m = 13 n  B ]
23. Q.12 Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is : (A*) 22222200 (B) 11111100 (C) 55555500 (D) 20333280 [Hint: 4 × 30 [x] + 3 × 20 [x] + 2 × 10 [x] where [x] = 1 + 10 + 102 + 103 + 104 + 105 Total number = ! 2 ! 3 ! 6 = 60 ; 4 remains at all the places 30 times |||ly 3 remains at all the places 20 times. ] Q.13 The number of ways in which we can choose 6 chocolates out of 8 different brands available in the market is :(Choclateof the same brand to be identical) (A*) 13C6 (B) 13C8 (C) 86 (D) none [Hint: consider 8 different brands to be beggar and compute the distribution of 6 identical things among 8 people each receiving none, one or more.Alternativelyfind co-efficient of x6 in (1 + x + x2 +..... )8] Q.14 An ice cream parlour has ice creams in eight different varieties . Number of ways of choosing 3 ice creams takingatleast two ice creams of the same variety,is : (A) 56 (B*) 64 (C) 100 (D) none (Assumethat icecreamsofthesamevarietyareidentical &availablein unlimited supply) [Hint: 10C3  8C3 = 120  56 = 64 ] Q.15 Number of cyphers at the end of 2002C1001 is (A) 0 (B*) 1 (C) 2 (D) 200 [Hint: 2002C1001 = )! 1001 ( )! 1001 ( )! 2002 ( no. of zeros in (2002)! are 400 + 80 + 16 + 3 = 499 no. of zeroes in (1001 !)2 = 2(200 + 40 + 8 + 1) = 498 Hence no. of zeroes is 2 ) ! 1001 ( )! 2002 ( = 1 ] Q.16 There are 12 books on Algebra and Calculus in our library , the books of the same subject being different .Ifthenumberofselectionseachofwhichconsists of3booksoneach topicisgreatest thenthe number ofbooks ofAlgebra andCalculus in thelibraryarerespectively: (A) 3 and 9 (B) 4 and 8 (C) 5 and 7 (D*) 6 and 6 [Hint: = xC3 · 12 – xC3 ] Q.17 Numberofwaysinwhichallthelettersoftheword"ALASKA"canbearrangedinacircledistinguishing betweentheclockwiseandanticlockwisearrangement, is : (A) 60 (B) 40 (C*) 20 (D) none of these [Hint: Consider A's to be different  Number of ways (6  1) ! = 120 the A1, A2 & A3 can be arranged in 3 ! ways  120 6 = 20 ] Q.18 Numberofnumbers divisible by25 that can be formed using only the digits 1, 2, 3, 4, 5 & 0 taken five at a time is (A) 2 (B) 32 (C*) 42 (D) 52
24. [Hint: 1, 2, 3, 4, 5, 0 A number divisible by25 if the last two digits are 25 or 50 (i) Now, if 5 is not takenthen number of numbers = 0 (ii) If 0 is not taken = 6 (iii) if 2 is not taken then = 6 and (iv) If 1 | 3 | 4 is rejected then in eachcase we have, = 6 = 4 2 . 2 1 if 3, 4, or 1 is not taken then 10 number for each case. Total = 30 Hence total = 30 + 6 + 6 = 42 ] Q.19 A libraryhas 'a' copies of one title, 'b' copies each of two titles, 'c' copies each of three titles and single copyof 'd' title . The numberof ways in which the books can be arranged in a row is : (A) ( ) ! ! ! ! a b c d a b c    (B) ( ) ! ! ( !) ( !) a b c d a b c 2 3 (C*) ( )! ! ( !) ( !) a b c d a b c    2 3 2 3 (D) none of these [ YG '2000 Tier II ] Q.20 Threedigitnumbersinwhichthemiddleoneis aperfectsquareareformed usingthedigits 1to9. Their sumis: (A*) 134055 (B) 270540 (C) 170055 (D) none of these [Hint: Middle place 1, 4 & 9 Two terminal positions 1, 2, ...... , 9 Hence total numbers = 9 . 9 . 3 = 243 (Terminal digits in 9 ways and middle onein 3ways) For the middle place 1, 4 & 9 will come 81 times  sum = 81  10 (1 + 4 + 9)  A For units place eachdigit from 1 to 9 will appear 27 times  sum = 27 (1 + 2 + ...... + 9)  B For hundreath's place, similarly sum = 27  100 (1 + 2 + ...... + 9)  C A+ B + C gives the required sum ] Q.21 Number of eight digit numbers which can be formed using the digits 1, 3, 4, 5, 6, 7, 8, 9 (without repetition) if eachnumberhas to be divisibleby275 is (A) 12 (B) 24 (C*) 36 (D) 72 [Hint: Anumberwill be divisible by275ifit isdivisible by25 and 11 . For divisible by x + y = 1 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 43 For divisible by25 last two digits must be 7 5 Hence Now we have 1, 3, 4, 6, 8, 9 to be filled in six places, such that sum of even places = 27  5 = 22 or 16  5 = 11 sum of odd places = 27 5 = 22 or 16  7 = 9 2nd option is possible when we have 1, 4, 6 at even places and 3, 8, 9 at odd places ]
25. Q.22 Aletterlockconsistsofthreeringseachmarkedwithfifteendifferentletters.It isfoundthatamancould openthelockonlyafterhemakeshalfthenumber ofpossibleunsuccessful attempts to openthelock.If each attempt takes 10 secs. thetime he must have spent is not less than : (A*) 4 1 2 hours (B) 5 1 2 hours (C) 6 1 4 hours (D) 9 hours Q.23 A guardian with 6 wards wishes everyone of them to study either Law or Medicine or Engineering. Numberofwaysinwhichhecanmakeuphismindwithregardto theeducationofhiswardsifeveryone of them be fit for anyof the branches to study, and atleast onechild is to besent in each disciplineis : (A) 120 (B) 216 (C) 729 (D*) 540 [Hint: Divide 6 children into groups as 123, 411 ot 222 Now total = ! 3 ! 2 ! 1 ! 6 3! + ! 2 ! 1 ! 1 ! 4 ! 6 3! + ! 3 ! 2 ! 2 ! 2 ! 3 . ! 6 = 360 + 90 + 90 = 540 ] Q.24 There are (p + q) different books on different topics in Mathematics. (p  q) If L= The numberof ways in which these books are distributed between two students X and Ysuch that X get p books andYgets q books. M =Thenumberofways inwhichthesebooksare distributedbetween twostudents Xand Ysuch that one of them gets p books and another gets q books. N = The number of ways in which these books are divided into two groups of p books and q books then, (A) L = M = N (B) L = 2M = 2N (C*) 2L = M = 2N (D) L = M = 2N Q.25 Coefficient of x2 y3 z4 in the expansion of (x + y+ z)9 is equal to (A*) the number of ways in which 9 things of which 2 alike of one kind, 3 alike of 2nd kind, and 4 alike of 3rd kind can be arranged. (B)the numberofways inwhich9identicalthings can bedistributed in 3 personseach receivingatleast twothings. (C) the numberof ways in which9 identical things can bedistributed in 3 persons each receivingnone one or more. (D) none of these [Hint: Consider (x + y + z)9 = 9Cr x9 – r (y + z)r = 9Cr x9 – r · rCp yr – p zp [x + (y + z)]9 put r = 7 ; p = 4 or 9C2 · 7C3 · 4C4 = ! 4 ! 3 ! 2 ! 9 ] Q.26 There are 6 boxes numbered 1, 2, ..... 6. Each box is to be filled up either with a red or a green ball in suchawaythat atleast1box containsagreen ball andtheboxes containinggreen ballsareconsecutive. The total numberof ways inwhich this can bedone, is (A*) 21 (B) 33 (C) 60 (D) 6 [Hint [11-12-2005, PQRS] allsixgreen 1 5 green 2 4 green 3 3 green 4 2 green 5 1 green 6 ]
26. Q.27 12normaldicearethrownonce.Thenumberofwaysin whicheach ofthevalues 2, 3,4,5 and 6 occurs exactlytwiceis: [ 1,1, 2,2, 3,3, 4,4, 5,5, 6,6 can come in any order ] (A) ( ) ! 12 6 (B) ( ) ! . ! 12 2 6 6 (C*) ( ) ! 12 26 (D) none Q.28 In anexamination, the marks forphysics and chemistrypapers are 25 each, whereas maximum marks formaths paperis 100. Thenumber ofways in which a student can score50 marks is : [assumethat the marks areawardedin non-negative integral values] (A) 716 (B*) 676 (C) 605 (D) none of these [Hint: coefficient of x50 in (1 + x + x2 + ...... x25)2 (1 + x + x2 + ...... + x100) Q.29 For agamein whichtwo partnersoppose two other partners, 8men are available. Ifeverypossiblepair must playwith everyotherpair,the number of games played is (A) 8C2 . 6C2 (B) 8C2 . 6C2 . 2 (C*) 8C4 . 3 (D) none [Hint: For each game 4 persons are needed . Hence select 4 from 8 in 8C4 way . Now from each selection 3 games can be had  8C4 . 3 = 210 ] Q.30 SetAconsists of 4 distinct elements & the set B consists of 5 distinct elements . Number of mapping defined from A Bwhich are not injectiveis _____. (A) 600 (B) 550 (C*) 505 (D) none [Hint: 54  5! = 505 ] Q.31 Numberof7digit numbers the sum of whose digits is 61 is : (A) 12 (B) 24 (C*) 28 (D) none [Hint: only 7, 8 and 9 can be used] Choose the correct alternatives (More than one are correct): Q.32 Identifythecorrect statement(s) . (A) Number of naughts standing at the end of is 30 . (B*)Atelegraph has 10 arms and each arm is capable of9 distinct positions excluding the position of rest. The number of signals that can be transmitted is 1010  1 . (C*) Numberofnumbers greater than 4 lacs which can beformed byusing onlythe digits 0, 2, 2, 4, 4 and 5 is 90. (D) In a table tennis tournament, everyplayer plays with everyother player. If the number of games played is 5050then the number of players inthe tournament is 100. Q.33 n + 1C6 + nC4 > n + 2C5  nC5 for all ' n ' greater than : (A) 8 (B*) 9 (C*) 10 (D*) 11 Q.34 Thenumber ofways inwhich 200different things canbedividedinto groups of 100pairs is : (A) 200 2100 ! (B*) 101 2       102 2       103 2       .... 200 2       (C*) 200 2 100 100 ! ( ) ! (D*) (1 . 3 . 5..199) Q.35 The continued product, 2 . 6 . 10 . 14 ...... to n factors is equal to : (A) 2nCn (B*) 2nPn (C*) (n + 1) (n + 2) (n + 3) ...... (n + n) (D) none
27. Q.36 The Number of ways in which five different books to be distributed among 3 persons so that each person gets at least one book, is equal to the number ofways in which (A) 5persons are allotted 3 different residential flats so that and each person is alloted at most one flat and no two persons are alloted the same flat. (B*)numberofparallelograms(someofwhichmaybeoverlapping)formedbyonesetof6parallellines and otherset of 5 parallel lines that goes in other direction. (C*) 5 different toys are to be distributed among 3 children, so that each child gets at least one toy. (D*)3mathematicsprofessorsareassignedfivedifferentlecturerstobedelivered,sothateachprofessor gets at least one lecturer. [Hint: Given answer is 150 which comes in BCD inA, it is 5C3 · 3! = 60] Q.37 The maximum numberofpermutations of2n letters in whichthereare onlya's&b's, taken all at a time isgivenby: (A*) 2nCn (B*) 2 1 6 2 10 3 4 6 1 4 2 . . ...... . n n n n    (C*) n n n n n n n n       1 1 2 2 3 3 4 4 2 1 1 2 . . . ...... . (D*)   2 1 3 5 2 3 2 1 n n n n . . . ...... ( ) ( ) !   Q.38 Number of ways in which 3 numbers inA.P. can be selected from 1, 2, 3, ...... n is : (A) n        1 2 2 if n is even (B)   n n  2 4 if n is odd (C*)   n1 4 2 if n is odd (D*)   n n  2 4 if n is even [Hint: n =2m, arrange the numbersinto disjoint sets 1, 3, 5, .................... (2m – 1) m number 2, 4, 6, .....................2m mnumbers no. of AP's = mC2 + mC2 ] Q.39 A man is at the origin onthe x-axis and takes a unit step either to the left orto the right .He stops after 5 steps or if he reaches 3 or  2. Number of ways in which he (A*) reaches  2 is 3 (B*) reaches 3 is 4 (C) stop exactlyafter walking 5 steps is 12 (D*) can perform the experiment is 20 [Hint: for C the correct is 16 ] Q.40 Consider the expansion, (a1 + a2 + a3 + ....... + ap)n where n  N and n  p. The correct statements are (A*) number of different terms in the expansion is, n +p  1C n (B) co-efficient of anyterm in which none of the variables a1 , a2 , ...... , ap occur more than once is 'n' (C*) co-efficient ofanyterm inwhich none ofthe variables a1 , a2 , ......, ap occurmore than once is n! (D*) Number of terms in which none of the variables a1 , a2 , ...... , ap occur more than once is p n       .
28. Q.41 Whichofthefollowingstatements arecorrect? (A*) Number of words that can be formed with 6 onlyofthe letters of the word "CENTRIFUGAL" if each word must contain all the vowels is 3 · 7! (B*) There are15 balls of whichsome are white andthe rest black. Ifthe number of ways in which the balls can bearranged in a row, is maximum than the number of white balls must be equal to 7 or 8. Assume balls of the same colour to be alike. (C)Thereare12things,4alikeofonekind, 5alikeand ofanotherkindand the rest areall different.The total numberofcombinations is 240. (D*) Numberof selections that can be made of 6 letters from the word "COMMITTEE" is 35. [Sol. (A) 11 c 7 v 4 /  7C2 · 6! = 3 · 7 · 6! = 3 · 7! (B) No. of ways = )! r 15 ( ! r ! 15  = 15cv      r W ..... W W      r 15 B ...... B B B  This is maximum if r =7 or 8 (C) Total no. of combinations = 5 · 6 · 2 – 1 = 240 –1 = 239 (D) 2 alike + 2 other alike + 2 other different = 1 2 alike + 2 other alike + 2 different = 3C2 · 4C2 = 18 2 alike + 4 different =3C1 · 5C1 = 15 All6different = 1 —— = 35Ans. ] Subjective: Q.42 On the normal chess board as shown, I1 & I2 are two insects which starts moving towardseachother.Each insect moving with the same constant speed . Insect I1 can move only to the right or upward along thelineswhiletheinsectI2 canmoveonlytotheleftordownwardalong the lines of the chess board. Find the total number of ways the twoinsects canmeet at same point during their trip. [Hint: (8C0 . 8C0) + (8C1 . 8C1) + .... + (8C8 . 8C8) = 16C8) = 12870 ] Q.43 How manynumbers gretater than 1000 can be formed from the digits 112340 taken 4 at a time. [Ans: 159 ] Q.44 If N = 2p  1 . (2p  1), where 2p  1is aprime, then find the sum ofthedivisors of Nexpressed in terms of N. [ Ans. 2 N ] [Hint: Let 2p  1 = q (prime)  N = 2p  1 × q also 2N = 2p (2p – 1)  Sum = (20 + 21 + 22 + ..... + 2p  1) (q0 + q1) = 2 1 2 1 p         (1 + 2p  1) = 2p (2p  1) = 2 N ]
29. Q.45171/1 Tom has15 ping-pongballs each uniquelynumbered from 1to 15. Healso has a red box,a blue box, and a green box. (a) How manyways canTom place the 15 distinct balls into the three boxes so that no box is empty? (b) Suppose now that Tom has placed 5 ping-pong balls in each box. How many ways can he choose 5 balls from the three boxes so that he chooses at least one from each box? [Sol.(a)315 – [3C2 + 3C1(215 – 2) = 315 – 3 · 215 + 3 Ans. (b) The 5balls canbe choseneither as 1, 1, 3 (1 from abox, 1 from anotherbox, 3 from remainingbox) or as 1,2, 2. Thereare3 ways to select as 1, 1, 3 (take the 3 balls from red or 3 from blue or 3from green). Thereare3 ways to select as 1, 2, 2. Thus, recallingthat theballs are uniquelynumbered,the answer is 3 · 5C1 · 5C2 · 5C2 + 3 · 5C1 · 5C1 · 5C3 B1 B2..................B15 = 1500 + 750 R B G 1 2 2 1 1 3 = 2250 ] Q.46 Findthenumberofways inwhich12identical coinscan bedistributed in6 different purses, ifnot more than 3 & not less than 1 coin goes in each purse. [Hint: 000000 remaining      (i) 2 coins in each of 3 purses = 6C3 (selecting 3 purses from 6 different purses = 20 . (ii) 2 coins in one + 1 coin in 4 purses = 6C1 . 5C4 = 30 (iii) 2 coins in each of two purses + 1 coin in each of two purses = 6C2 . 4C2 = 90 (iv) 1 coin in each of 6 purses = 6C6 = 1. Alternatively : co-efficient of x12 in (x + x2 + x3)6 = 141 ] Q.47 Duringthetimesofriots, residents ofabuilding decideto guard their buildingfromground and terrace level ; onemanposted at eachofthe fourdifferent sides and onewatchingthe compound gate.Ifout of 11volunteers,2sufferfrom acrophobiaandother 3wish towatchonlythecompound gatethen find the numberof ways in which thewatch teams which can be posted. [Ans. 25920 ] [Hint: one person for the compound gate can be taken in 3C1 ways and cna be arranged only in one way 2 positions on the ground can be selected in 4C2 ways and two people can be arranged in 3 ! ways 6 persons for theremainingpositions can be filled in 6!ways . Hence total = 3C1 . 4C2 . 2 ! . 6 ! = 25920 ]
30. Q.48186/1 10 identical ball are distributed in 5 different boxes kept in a row and labledA, B, C, D and E. Find the nnumber of ways inwhich the ball can be distributed in theboxes if no two adjacent boxes remain empty. [Ans. 771 ways] [Sol. Case-1 : When nobox remains empty itisequivalentlydistributing 10 coins in 5 beggar           4 5 Ø Ø Ø Ø 0 0 0 0 0 = 9C4 = 126 Case-2: Exactlyoneis empty 5C1 Ø Ø Ø 0 0 0 0 0 0 6      = 5 · 9C3 = 420 Case-3: Exactlytworemainsempty ( 5C2 – two adjacent) 9C2 Ø Ø 0 0 0 0 0 0 0 7      (10 – 4) × 9C2 6 × 36 = 216 Case-4 : Exactlythree empty. There is only1 wayto select 3 if no two adjacent Hence 1 · 9C1 = 9 0 0 0 0 0 0 0 0 Ø Total = 771 ways ] Q.49 Findthenumberoffourdigitnumbers,whichcan beformed usingthedigits of thetwelvedigitnumber 123456432424. [Ans: 779] [Sol. Thegiventwelvedigit numberis 123456432424. It has four 4’s ; three 2’s ; two 3’s and one 1, 5, 6 each. Toform afourdigit numberusingthedigits ofthis number, therearefivecases possible. Thefourdigitnumber canhave (i) allfourdigits same. (ii) threedigit sameandonedifferent. (iii) two digits are same and the rest two are also alike (of other kind). (iv) two digits are same and two are different (v) allthefourdigits aredifferent. Numberoffourdigitnumbersformedwith Case (i) is 1 Case (ii) is 2 × 5C1 × ! 3 ! 4 = 40 Case (iii) is 3C2 × ! 2 ! 2 ! 4  = 18 Case (iv) is 1. 5C2 . ! 2 ! 4 + 1. 5C2 ! 2 ! 4 + 1. 5C2 ! 2 ! 4 = 360 Case (v) is 6P4 = 360 Thus, the total number of four digit numbers = 1 + 40 + 18 + 360 + 360 = 779Ans ]
31. Q.50 Findthenumberofquadrilaterals whichcan beconstructedbyjoiningthevertices ofaconvex polygon of 20 sides if none of the sideof the polygon is also the sideof the quadrilateral. [Ans. 17C4  15C2 ; 15 3 20 4 C . ; 2275] [ Hint : For (B) Select saythe vertex 1 of polygon . Now 2 & 20 cannot be taken. The vertices remaining are, 3, 4, 5, ......, 17, 18, 19 . 3 more are to be taken which are say o o o. Now we have 14 crosses left. x x x x ...... x x. Then there are 15 gap and 3 can be taken in 15C3 ways . The total ways = 20 x 15C3 , but the same quad. will appear 4 times. Hence the required answer is 15 3 20 4 C . . For (A) 1 2 3 4 5 ...... 19 20 As in linear, out of 20 vertices (1 – 20) 4 are to be taken i.e. O O O O (to be selected) X X X .... X (16) not to be taken out of 17 gaps 4 can be taken in 17C4 ways Hence, n(R) = 17C4 –number of ways in which the vertices (1 and 20) and 2 other nonconsecutive can be selected as 1 and 20 becomes consecutive vertices in circular. now, 1 and 20 selected, hence 2 and 19 connot be selected. Remaining 2 more non consecutive are to be selected from 3, 4, 5, .... 17, 18 (16 vertices) O O (to be selected) X X X ... X X (not to be selected) (14) now out of 15 gaps 2 can be selected in 15C2 ways. Hence these cases must be subtracted.  n (R) = 17C4 – 15C2 ]