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Main 4 ane Gree in Adv Y Expected pULfed fe Pasbely i ae Phe Pocus - FUNDAMENTAL PRINCIPLE “ COUNTING | oF (tach enarbie hes 3 options) a6 SS * ry 000000 Beeps beat : SOE: ng Rr aakee Total ways = 35 Sa Se Su Sm 5 = 3* inf Q. Total 7 person gathered in conference hall for Annual meet, find total number of Hand-shakes G@) Total ways = (xyz) ways Sol PRs) Pa Pas Paehase ae Total handshakes = 6 + 5244302 e2=22 ROS* Bus “mw | rite ido There are 7 flags of different colour. Find the number of different signals that can be “Addition raga (OR) ‘Agr giver. Atsss a? pe eigen Sa complete by ae transmitted by the use of 2 flags one above the other f i Br cess process ko complete _krne ke ‘n' aur no. tsipal ames #2 signals fnf. Q. Section-B of JEE-Adv. contains six multiple choice questions in which one or more than one options can be correct. Find total number of ways to attempt these & questions: (i) Each question should be answered. (a) Onecan leave the question unattempted. | Sol (i) Each question should be answered. qa 1a hed i 36 415 15.~.35=(25)* (i) Any questioncan remainun-attempted Total ways = (16). Q. A boy has & marbles, number of ways in Which he can put these marbles in 3 pockets. (id) IF the sum of digits is divisible by >. Oven number is divisible by 3. a iid) Divisible by ‘4! = Last two digits of number is divisible by # (iv) IF the last digit of the number is 0 or 5. the number is divisible by 5. (¥) Divisible by “6 => Number should be divisible by 2 & 3. (vi) if the sum of digits is divisible by 4, then the number is divisible by 4. (vil) Divisible by 20' = Last digit of number is zéro. (vill) A number will be divisible by 11 if the difference of sum of digits at even and odd places is divisible by 33. (0) Whose last two digits are divisible by 25: (*) if last three digits are divisible by 8 => Number is divisible by 8. (ci) For divisibility by 7, double units place digit & subtract it from rest of the number. ()) abe — “divisible by 3* fabe =a *300+b» 507054 = at9 + i)+ 6 (9+ t)eeu d => 99a + 9b +(a+ & +6) diy, by 3. sum of digits ~ div. by 3 (ii) abede -> “divisible by 12" abide = as 20+ be 1 ew 1 +dulowe => a(9999 + 1)+ W001 ~2)+ (9942) +d3t- thee divisible by 12 always 99994 + 1001b + 99¢ 4 144 Sal. abeabe = 300000 a+ 20000 § * 100, ‘i és = (100100) a + ie died b+ (2003), 7 £1x91x100 L1n9ixt9 242. = 91 (1000+ 10b+¢) sii x 13 7 (abc) => Hence (abeabe) is div. by all (12, i> Q. Six digit number is formed using only sgt | 8 is always div. by _ fatten aug. oo 5 TRAL= 2 hen yy gaeiea8 Tas * 2= 8 whens of Sale aeaee@8 jo Pup ; http 2a [ts +e, ae C= Mey tide mg t- 7+ fame wind! = (0, 2) abedef = 200000a + 100008 + 10006 + 1004 + x04) |) = TIGR + LOIN, + 107K, + 10, + 3 oh H30% + 308 + 10+ 30% 303g | = Wint) + (2242.11) > 121111 also div, by 7 . Number is divisible by 7 fnf Liaris RMATION F NUMBERS > UNDERSTANDING MEANIN I PERMUTATION & COMBINATION [IEEE Note-01 3 = as , Total = Exactly 'O' + Exactly “1° » v Let ¢lass has 4 students : A, B,C, Di | asia tat on aty ()) Call any 3 studonts on the taleetion) oR + was = @ DB @)ieminans mz . Call any 3 students on the ground. de - 2: ’ (i) iain We een ote Atleast one Total - (zero) At least two = Total — (zero + one) és) & @ @ { At most one = One + zero At most two = two + one + zero Note-O3 Bijection Method: Required = Total -— (Not required) Q. A dice is rolled 3 times, top-most faced | considered as outcome then fina the numb of outcome such that: () 5 appears atleast once (i) All 3 outcomes are different agc ||\aep Ace# |/ace8 SAE ||840 eca |\soa cas cas cna ||O84 (iid) Ex 2 1,2,43 (5, 5 o => Select 2 No. = "Cy (say: 3 de 4 selected) -5 select which will repeat (ways) (3, 3, 4) 52ers 4, 4) 3, 4.4 = Arrange them = 344,35. 4 4,4,% fof=*C,*2*3=490 TRICTED SELECTION i person are always or never @ Sboys are selected out of 4 boys such that: ()) Two “particular” boys will not serves together. (i) Two specific boys will only serves ir. Sol (i) "ek saath TH aayege” “ x x xx a4, 9+AB +A B f= (70, + 7C, + 7C5) (i) "ek saath Ai aayege" wv «Kx casts; AB + A B fa = (70, + 705) & Total s different need to select os 4 shoes such that () No restriction Exactly 1 pair should be formed (ii) Exactly 2 pair should be formed (™¥) No pair should be formed. paisabnceiand ire you "nettion & Combination Sol. (i) Ne restriction = °C, (i) Exactly 1 pair chould be formed | select i par | select 2 thoes SAC, (MC, + 96, + 8E, x 7C,] LL RR L DAR (iil) Exaetly 2 pair of should be formed ="Cy (iv) No pair should form faa bi kykghy No. of ways = 30,9C;) byl fof. = (90, +4C,+ 026, +7C,°C,* BREESE sijeceion = Reg, = Total = (exactly 2 paar + exactly 2 parr) =30C, - (70, (00, 40, « 1€9C,) + *C,) | Question | [JEE (Adv.)-2016] Q. A debate club consists of & girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these + members) for the team if the team has to included at mort one boy, then the nunaber of ways of selecting the them is (A) 380 (8) 320 (Cc) 260 (Dyas Sol. 6G + 48, 4 mem. to select in which 2 caption cases: (10 3G) + (OB 40) = 40, 016 804 CCl xt =404520+41554=320+60=380 inf EET (ce (Adv)-2018] Q. In a high school, a committee has to be formed from a group of & boys M,.M., M,, M,. MM, and 5 girls G,,G,, Gy, Gy. Go (i) Let a, be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls. i) Let o, be the total number of ways in Md Shien ieiigrconlonitban aan ie: foreked such that the committee has at least 2 members and an eqn nub obs nae < poo @. There ave + straight lines, % circles and 2 | parabola's are present, then find maximurn ber of point of intersection possible sol b-L ¢-¢ L-p Mat POH AC, © LCC, x oe “CPC, 2 &-Pp P-p +O, = 4 +40. 4 I -oUNTING RECTANGLES H & : SQUARES Sol, Let: (m= 5 den = 4) = Sheet: (S* 7) Horizontal Length of & unit = $ Length of 3 unit = 3 Length of $ unit! = 1 | Vertical Length of L unit= 7 Length of S unit = 5 Length of 5 unet = 3 | Lengths of 7 unit =5 fPef="c. waee = orc, = WTIE METHOD h & GAP METHOD ff Ble wae (KK PMOC) Wa we) oe ee rere rt (All are together) TIE METHOD use Q; Count total possible squares and rectangle | Q. Arrange a, b,c, d he @ in a line such that in given grid of (@ « 7) Sol. Count squares dimension wise Squares of 1 x2 = 7x G = 42 Squares of 2 « 2 = @ x S = 30-7 an Squares of 4 x 4=4 = 3522} tH Squares of 3x 3 = 5x4=20/4+ Squares of # x4 =4 «3 = 22 ++} aa Squares of Sk S 23x 2=6G Grid(e« 7) Squares of xG = Ze =2 add Now, Rectangles: Total vertical lines = 7 Total horizontal lines = 8 So, Total rectangles = (7C, « *C,) inf. [llT-JEE-2005] & There is a r sheet of dimension eeest) = x (2n — 1), (where m > 0, n> 0). Ce eee ines seen ink ace Fed the sdee of octane hats 200 oF odd unit length. Aims ns sj (By mnfm + tYn + 4) (C) 4mm _ (D) mn? Permatation te Combination — a, b & ¢ are always together, Sol. Number of ways = 3! = 3! vt str (inert arrangement of 3 units of a, b & c. labeldve Tie method: (yaa witet wr erat em internal arrangement 1 Al) re OT al a BT AT FT) (No two are together) use: )GAP METHOD Q. Arrange (a,b,c, d,¢ & F) ina line such that no two of a, b & ¢ are together. Sol. Jinko 4em sem Tan f sre GAP A ran ay Se ea oe ine sees! =a! (Selection a oae Seopa elisa e&f) ai (i) All are together (abcd) = TIE Method (li) No two are together (abcd) = GAP Method (i) All are not together = BIJECTION (abe_d) OR (ab_cd) or (ab.c.d)—. Q. 4 & 4 girls are to be seated im a line and find number of wayt (ii) iF not all the girts are together Or if at least one girl G@ teparated from vest of girls. (ii) Boys and girls are alternate (iv) If there are 4 murred couplet then nuraber of ways in which they can be seated so that each coupla & together. (v) /8,8 G.G, 8, B,. G,, G, >. By. By. Gy = Gig = 2 (70, «2! aha 2! 0 4) inf (W) Bosse Gy. ew at By -nan Ga OH aal 6, . _G, ~~ 6iea2! inf = Se fe dew yg 8 & G ~ alternate (BG OG 8G BG) + (GB G8 Ge ce (we) B,G, \-* 7, «Sees } i fal = 20 Sf dt = "C, Ble 3! fH ARRANGEMENT WHEN AL ARE NOT DIFFE of A, B.C, DLE (@) Number of " (takin all at a time) = 57 (Number of arrangements of A, As A, 8.6 RANK OF A WORD HIN DICTIONARY When all letters are different. a Q. Find rank of word GANIT Sol. A, Gy l.N,T Q. Find rank of word MASOOM Sol Zot #55 ae M A SOOM Ans. =- 4 GAT 2! Q. iP the letters of the word ‘MOTHER’ GANIT=3 permuted and all the words so formed (with or without meaning) be fisted as in a dictionary, then the position of the word Rank = 24+ 2+3 = 27 inf Case-tl When ail letters are not different. ‘MOTHER! (5 __ So E HM ORT | Example | Ee nth 33 Q. Find rank of word MASOOM Mm aie Sol. A. M,M.0, 0,5 are =I ME___ _ =» 4! A----- cory es Mee a MAM > 3 3 MOE __ 8 st MAO ___~B3/=6 MOH = == 3! MAS M. > t=% MOR __. + 3H MASOMO—-1=2 MOTE ____ at MASOOM—+2=3 MOTHER + 4 Rank = 42 Q. Find rank of word GANIT Soi BR) Race: AG itn T Rot o 2 8.0 Ryo sia st of Rank = 354 +0%3! ed eter onde Ones = a7 ff = 240 nica +1f+3=309 WE «3! = &, count = sto R ae euMa=-~ 3! = 6, count = s1¢ HOE ~ 2, count = Sig R - bie fano MET» 1, count <5 so¢ ——— 3 ial the letters of the word PUBLIC are : ted and words thus formed are | ranged as according ta dictionan iothen r at 457° paces y word written Position js sep ell Pu g...-~ Sl = 120 —— 2512120 ~ S! = 220, count = 360 -_ 7 #! = 24, count = s¢4 i LPS C__— 2!, count = 434 + 4! = 24, count = 4352. LPS) + 2!, count = 435—6 fal LP | > 2! count = 432 GREE vee Main-2023] Q tet S={1, 2. 3, 5, 7} The rank of 35773 if all $ digit number formed by the set S are arranged in a dictionary in ascending order and repetition of digits is allowed. ieee at 35. 5_ 7 5 ae st Se oss 5t_lii+s? 3S72_35 320.353 SS73_35 a 5 S575 2—5 Tet os 52 BS771—-2 Rae by ca 3577242 Ej paeerer | 3S77T3~2 Total = 1748 inf I cose ccin | Question | [JEE Main-2023) Q. iF all the x digit numbers My Mg Ny Mg My *, with © « x, eu, Ox, Ow, tm, "Ce tos — 2 *C, «i245 So, 72% number ig} 2456748) Sum=2+4+5¢6+7s8= Number of ways of selecting three letters out of A,B,C, DE (all different) > *C, = 10 ways a ‘ Number of ways of selecting three letters out 444, 85.C, D Sol. Case-t: All alike (AAA) = no. of selections = 2 Case-thi 2 alike, 2 diff. (AAB, BBC, AAD.) = no. of selections = 20, « 9C, = & Case-itl; All different (ABC, ABD, BCD...) = ho. of selections = 4¢, = 4 ofnfict+e+¢= at Q. How many 5 lettered words can formed using the letters of the "INDEPENDENCE". be & iF —_—— OS ee ae books = atheart 1. bape Laeort’ § book) sper e* ?—4)- (7, = te 3) _gateast 2 books of each subject iY, “afatleast2 Physics)» (atleast2 Chemistry) a (at least 2 Maths) eet (Cn ° *C)) x (S48) ayn ee 3) - 2) “s(@ -5)*4+* 5 = 220, ‘(a Baaetly 2 books eM, ebed + *C,2+ O24 (Exactly 2 books of each subject Paifiutx=3 Prime factorisation divisors = (© + 2)(B + 2). proper divisors = (x + 2).(B + 2). 2). <2 the divisors = (2° + 2 + 22 6... PSP e+ BAYS + Ste Le Rt than So Neate (i) Total number of divisors winich are Perfect cube w (96 Sy Rs Bee (iv) Total number of divisors aivisiivie bay 8 a(7 + soya + thet eee (V) Total number of divisors divivibte by 19 | = (7+ te «2) j (vi) Total mumsber of divisors divivlble iy 22 téues (2st) 1 (vii) Sum of all divisors = (2% + 28+ 2 ee 4 #27) (Bo Bt o_o BH (gs gs FY (viii) Summ of all “ever diviuars® © (2" © 2” « Bele 20 (Se Bee 4 ( i + Ss st) (i) Sum of all proper oad divitors = (3° « S++ BY) (5? + ste = 8 Sum of all divitors divisible by LO w (24+ 2ke io a 0 (Po So + 3*) (5* © 5°) (x) Number of ways in which "N' can be resolved as product of two numbers ae7«3 (x Se (xii) Number of ways in which 'N’ cam be resolved as product of twa co-prime numbers = 29° = 4 (xiii) Number of divisors of the forw (4) =4ke multiple = 60 Te 3 = 126 (xiv) Number of divisors of the form (4a + 2) 1 = aise divisors Jinko 4 se divide krne i pr remainder = 2 ap 4n¢2>2(an+ 4) = 2 = (odd no) = No. of di. tis Ta Bean (i) Munober of tices eter item) #2k # Ss ki powers: (4n 42) 8(4n et) “(Cn #4) Divisor = (4n ¢ 1)('4n¢ 2) (40+ 1) adn + 3 form No. of divisors = 2 «4, « °C, * 42. fh 4n 7 126 eT i> 12 ie 168 se dy » 2 + 24 n+ 3-4 | Question | [JEE Main-2023] Q. Let n be a non-negative integer. Then the number of divisors of the form “4n + 2" of the number (10)'°(11)4.(13)** 6 equal to N= 290 570.4413 4555 = ram 2) 27.2 (4% +3) Cano 2) *(4n) = 5? st 5%. 510 Sol, (4m +1) = all 21 powers of 5. = 12° 113 217 113_ 334% (4n + 1) form + 6 (12°, 117, 114, ..., 2179) (4n + 3) form + 6 (114, 119, 117, .,11"*) = 23° 1353 735)° all 12 powers of 23 - (4n +1) form Divisar = ( POE Gy eg een ) (4n + 2) Sy + aati £. ANC * sia +1) Lu 22 «© 6 w T4 SS es Rebates tesa i can be arranged on circle = 2. Number of ways in which ,_ iP things can be pees on creo clockwise & anti-clock-wise arya ; (nk are considered as same= ee Clockwise & Anticlockwise arrangement in considered as same in case of NECKLACE oy GARLAND (HIeT) or ALL HAVE DIFFEREW: NEIGHBOURS in every arrangement. 2 s Q. Find number of ways in which 58 ad 5G tan be seated on a circle alternatdly! particular B, and G, are never adjactth each other in any arrangement, Sol. SB, 5G Case-lt; When group size in same. | Sol. 6 books + "4 persons” 2 | Sunita: antag making © different NC SS ns 6! 2 C-;2,2 3 seen ert aieiks Call! 3) yt hee é! = No. of ways = | Diaiar “3! B/a/giaisi* 3 (C-1) + (C-il) = 1560 inf. 10 different_-> BEE Direct formula objects ~2 = 46 — (40, « B4) + 40, n 29 — 40, 0 (2) 2 = 1560 inf. = No. of ways = ren iP + * 4 [EE (Adv.)-2020] ixn3!}) 2! at Q. In a hotel, four rooms are available. six | a persons are to be accomodated in these 7 different —* 7 four rooms in such a way that each of objects —— rooms contains at one person and at mets most two persons. the number of all possible ways in which this can be done is_ # , TY eee R, R, Ry R, : = No. of ways | Syaiaisiz!) 2x3! mere) OO) @)© 2, 2, 2 epee 5 Jitne group ka size same hai utne ke factorial inal e! it se divide krna hail . Halatamaiat) "oo 3 N ¢ Number lumber of ways ' of dividing int oF ways of 4 Geiss into distribution these groups 4. Number of ways of distributing 'n'-different | Example | things among ‘n’-persons (such that each 2. Number of ways of distri ‘na 7. different Coalameoe i in'-different chocolates to four childrens such that each | 2, Number of ways of 4 gets at least one chocolate. things among 'r'-person (all possible ways) | soy | Sunita: e rnrar = (ry ways ae Cases: & Number of ways of P ‘w-different i acne on)-»{(cse making) °F Direct formula = —"C,(r — 49 +°CAr — 20 -Clr - 3h -_ .. SION & DISTRIBUTION oN DENTICAL OBJECTS angel of ways of distributing 13 identical @ things to 5 children such that: (No restriction (fardt at fart wh aay (id) each gets atleast one. sal () Consider 13 identical things: OO 0 oO ‘sik 4 identical sticks: 111 (23+ 4) number of arrangements = Eacharrangement is equivalent to a distribution ABCDE goo|joo|eceao|e00| = 3 2530 |ja0000000| ox C00O||=Oos SOO heeeecboiclooclo = #41431 (i) Each gets atleast one (sticks gap 4 Ta) MOO 00.10 Total gaps = 135 +2-2=12 NUMBER OF HINTEGRAL SOLUTIONS (BEGGAR'S METHOD) Eaton: Case-|: HF (x,, No Mays ons Be) EN Number of integral solutions = "~ Case-t: FRX x, ux) e W Number of integral solutions = "-*C,_, &. Find number of integral solutions of following equation with given conditions: () k+ y+2+t = 16 such that (x, y, 2, t € non-negative integers) (i) x+y+2+t= 16 such that (x2 4,y22, 223,t20) Permutation & Combination “Ss r-2 Sol. (iil) x+y +z+t= 16 such that (x2 4,y 2-2. Zz -3,t 2 —1) (Wx + y+zn+t © 16 such that (x, y, z, t € odd +ve integers) (v) x * y+ z+ 4w = 20 such that (x, y, 2, w € positive integers) (vi) x+y +z 20 such that (x,y,z € non- negative integers) (vil) 35 new eg: x +y' + 2+ t= 10 a Lé-t-1¢7 = 24 fnf = 24-30, = 4-1 (iil) w+ yez+t= =x' +P fof = 2480 oC (WW x+yez+t= 16 = (ap +3) (aq 3) + (are 3) + (ame d)= 26 => 2p+2q+2r+ 2m = 12 =p+q+r+m=6(p,g,r,m€ W) fnf = 4-40, _, = “Cy x+y+2+4wW=20 wt Lox+ey+z= le = *C, = 35¢, we Se ee ™ 1 {(s-y 2 P coeffofx*Pin || = “| x = Conff of * in (22 = (neyo, 0) case-t: (1) x? (77C,,) 4 (10 case-Il: (2) ¥° (°C, ,) xB? (—4 ys case-Wlt: (2) X° (°C) 07° (ayy Case-IV: (2) X° (PFC, .) xt (2)x9 fof. = (a oe - gc 4 so rs bia te) PRE Removal of extra cases Ge eRC. +c. = 30 #@when: C,24&C,22 0,4 ++ C,= 24 Total solution = 74°*-*C, | = 27¢. # When: C,2 8 2€,+C",+C,+C,=20 3 number of solution = 204-26, | = 23¢, # When: C,2 7 >, ¥C,+C',+C,= 19 number of solution = **4-2C, , = 22¢, @WhenC,2> 8 & C,27 erty e Ch + Cy +C,=15 number of solution = ****-30, , = 2*C, inf. = 270, — °C, + 22C,) + *8C, @. A train having 12 stations enroute has to be stopped at a 4 stations. Number of ways it can be stopped if no two of the stopping stations are consecutive. Soh *Susestsm where: x > 0; y,z,t2 2; M20 X*+y+zeteme s keyeztettem= s All possible solution = *-8C,_, = °C, ff. [EE (Adv)-2020] Dist Q. An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1-15 June 2021. is Q. Number of octagons (8 sided polygon) that can be formed by joining vertices of leosagon (20 sided polygon) such that no side of Octagon is common with polygon. Sol. => Select any 8 vertices = (such that no two are conseuctive) c x ella finf = Number of ways of derangement of n-things = D,, Q. Six cards and six envelopes are numbered 4, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card Find number of ways such that: (i) All cards ave in correct envelopes (ii) No card is placed in its corresponding envelope (ii) Exactly 3 cards in correct i ly 3 cards are placed in pes (iv) Exactly # cards wrongly placed Ee la cecal bala the card 1 is always placed in numbered 2. (JEE (Adv.)-2024) Sol. (i) All cards are in correct envelopes = 1 way (a soca is pei ee corresponding (iii) Exactly 3 cards are placed in correct envelopes = *C, x2 xD, = ‘op nix? (i) Exactly 4 cards wrongly placed =*C,xi=eD,= *C, x14 sic oO VI D => Derangement of 5 letters = D, = 44 = During derangement: Words starting with V = 12 Words starting with D = 12 = fnf. = 22 words. ‘ MW i i n(A wv B) = (A) + n(B) — n(A > B) n(AB)= (A) + n{B) +A(C) = (AB) —n(A0) -A(BOC)+ H(A Bo Cc) [JEE Main-2023] Q. The number of permutations of the digits 1, 2, 3, .. 7 without repetition, which neither contain the string 153 nor the string 24 67,is Sol. Let: A= numbers containing (153) B = numbers containing (2467) Total permutation = 7! G53),2, 4, 6,7 2a)= 51 G4e7] 1, 3. 5 => w(B) = 4) E53] Baer = wane = 2 = Req. ways = (Total px ‘ —_--=—=---, (v) During derangement of 6 cards, card number 1, 3, 4, 5, 6 will equally placed in envelope number 2. No. of ways = 285 - sways Q. The number 
