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(er pos) ster) a 6 | at SO) Qe a aioe ee | ‘un Tent "2ei-0" 2 [teat 4ste tn yh nan fensy) Lt SO) Z(ei-v*- n(2n- Doan $2 4 ss pom sip: ub Mel SQ) = 108) 4 oe gC) bun i || wy - Ludarh ve Shh 1 We Ouse Ahel Sn) vi Fee for Awe “ent & . SCH) = z Ot ore (EOE iaey b& a don Posh, y sta) - (kay (aio -1) o(t+1) +1)) of | . Ce Caa) Oe ar | 3 a kt st) = ERs -1 tot Hh sD, Gay We A frees) (Her) + 3 Gay) a | Coaifietonn) + sO) lr On Lt be+3) (2% rH) bea tsk ts) Gu oerireestey, Gen) 24 Ceres Gey) bree + 5) hn) +3) (An) shea) * | SU) We faeyp Mn né2t, 2) z nlne = po 2 SA Bhp EMln) Y a Ld sty) = Zany) = n (wan (tr) | qe 3 Basin soy Ld el 0 SC) = (9) (39 oa ye SC) en Bue 3 Tudeh ip! Assume Aorak Cup buns fot Some ext SS Ek kay 2 Klee 2) t=t 32 idee bee Hante 4g Pha (aay = Ga (ss) Cars) we browe te Ska = S(t) az brargde-¢4e (hay) 4 (eriplete) & = 2, kleti) 4 Er (e+) tel _ EC ey ey ete) a &by mothanolizol Dudek, prove tab nly tl 74 vn ezt SG)! nL yo" Ynez hoses steps JR M=1, SCi)> Ip at Nop of nee wt Benes os $07 ee Dunluctie Stp! Assume strat $4) a Hoe Jk some 4 £25 kly gt! a gt Pa = (i) fk steep, m bow be pt, oh SCL vs Pere oe — oedhing & 09 < (kn) #} ok kal {t 47) < (eat)! 1 2 a MK EHV K wis we (431% 0 Sta) ne HG pr ans me) for at postve satleg 7 sty: ane *-7) Loser Step: Lt nrhee6 S(a)=2 ss) e 46 <4 S6-7 = & 1 24 < 24 < SQe) vig truce Fader tp! sn) ban fe ea oe. uk 2 (Kr D —vV pe Drone te didn sta bo f sari) Yo nye white moné a gous LEZ bah 76: o., Ce 2 (&t-V— 4 Cha) eA Ce -7 Le) Gers) £ry Gath [becomes whine k7/6 , We boone Par e - = 1874 | ater 2 "7 + sa) we fre: — Der o> ent Hn nez* ood ME nD, 7 s@jrorpur ¥ nezt & n7s Bisa sup) LA ME 2% ‘ist eo s@)i V7 5* S (0) dee woe indus Sh s(x) fron fob Bome hen & bys cc): ke el? Re Wwe drone fe dads tu) Henk 7 a a" > (an kit A gels dea eee (aeeately ADD) “nigy? al oy S oe Ga het yeh tehe phy U4) , an gent raay* ys 5*7 2s) +) 7M Ben == [.. ke (aa)[SP fA Ez, 5 duds yon — Ld SU)! & bwoes nen Baas sip: Lt M21 807; 5 diwds 17-13 © olwds o. 1 SQ) ta Lee Duduckive 1 issue trate Stu) te Lew {* Age b eé 2+, S dwdes Fk Ss | 8, eR a mnt ge Meee | Tin oteas a fa mez? __G) Me Drow te hides tee Hate of C41) (KH) ¥ (esi) = (44 Skt 10k 410k 4k 41) — le ) | = (Ke- H+s (x4 4 10L54 lok? ask) gm + & (44 tob4 toes 54) — by 5 (wre kh 410k 4 fo eo 45%) <5) miter, Por te 4 108 fo Leash RY Tea Brows Bros Cena de a witha J 5. 9G) te Be ) Are “oui pork tabigi My 24 Ror Be Waitin os oa wen of 5% pudlor 97's. Son): Wn Lar be betittny AA Aun of SF lpr 7 ny 24 fosuls Shp: 5 (24) 245 (1 1) +(S +5) SSC oe Tee v vSurdarehs ve Astin Mook cCa) ia Brg kezgt £ ky 24> - be (le dae Da repens) Supface Te apelin. of A nahn ¥ se B eon & AFD rn if i aus i! Using Bir Bpetindads TA, we Prd tral kes (1874 + (Sree D4) | we Garr) C949) HET A 40 Gs i (Itt) +S rse- 9) Ces (7) (ATS) tact 1 Tha Slay teat SCk) wh a Aum of TA & sis. SCnp we Fem Cri le te fe Magis ad vet Risk ass Seneelaeeeenes + Ong ie hoa: bin an Oe vyonézt 18M) 1 <4" se Kass _sdy gy a1 Xa A225 3, 427 3849 * | neal? Lt SCay ws fea foe . feat S04) hy Aero gor 462" ie SY Asus. Fudnchrarm St! z | LAY 3 On <3 | MI 0, pO Mao | koe Oye aka kt 43 mr BN ge sh a ste 54)jus, sty) & Baa 7 Moy Mort, Mat? > 7 Ay Sodus Sep Arte tren SOD ae Heat fe at Netty ok Yo, Moat, ot*, ae ma pur £7), = 29: ‘ qh, sCk- ) @., sC20-4) & ‘ae 1 _ eon PC eee “s g¢en) <8 Pt , sas poe ¥ | sete deg etd sh a | ae bye we | | g Moor f 7% ws tg acd 5%} ofwe hel, Hed, ays le babe Hy, cla dade te! pn Tey eC Hae nex" nw ges 8G: EH HH tol fas Sips Lk nol ete 2H,71 2 axt-l =l sa) ws tem india pnt Tak SC | al fu cH = aot > Kal ie Hy ii \ ee | = (ke) Ha ei | 5 f oT aCe) | tye Gaol -# + Hee) al kl | oc GpNG1 7 ar ~ ke Hee] SF | sche) 2 he - (r+0). eS |= clap ve et ee (ies Et pow OD panna O44 5 Jatt +S fu ppt) 225 = 54Gt StS eS 0 eo fad T+ yease ete TA SAGAS 4ST | Wo +4 72s F qxeTtTeT7 | } © RECURS WE DE Fini ons Styruaces fh Keaprtnce vis Dw gf cmonds |g & dat Am Candin ab the finse Steels) re Hear 5 vo 6uyprintr, Wn tHAch On Eg: Consdi the snob. Stapalrrte be, bijsesy b,=2n fm out nen Hist, we fad aot : b, 2 2020 byt ls byt each bys be sre. sha fot uy othe WEN. (the Lupheit a by >on) - fi Sr bre Seapinen OI trnpbicrdly a ma he pond fame is suitedue wulo fe Fi Styne fom | Ae aca m a4 - My, bs bo, - be Wise baton % new ton iy a fatjng hie Stgytante oi he by +2 ae nzr anal b=2 >Ktwst dept - fie few dees Bee sequence pu Unditabiol vifscy Chace tee 6 woud poo_— Exawples a wbbdin OX AeUrs! vite. fr Lew Si {a, D |e lack yf te pris iow VC DO rH Wao “UPA, =3n+7 Wayanrtrts) yoann’ Wane 2- cy" a. Hee, a ee cblaiiest fs uF . ae ay-l | | N oa 3 a7 8%, wt | of | a eR ye" * 0 Fes | aslo, 213, A321 aye | | “a os ne +s | ~ noe W) ane nln” OF 3, A= %, O16, Oy = Ur Ay HESS UNAS, Agr A= T= 2XL*S, Ont = I= 243 +5, Os 4g all = 2X4 43 2 [Baar a= 23 | W+s {* nyl Z202> AN ESS 2K2+1, Ay-AR T= 2X3 tH, Ay-%=4=2K4 2 Pac Ont eet | pensyO ee DP Oyen A=}, aagr » %%24 1 ay 216 Qo -O, 28's 3 Pehl aa- Gass aka 41, aa axa Peace] fe vo ys Ou= 2-C-1)" AS, Aral, Ax 8, AF!) ages 451 : we’. wean, eee RE a all ie AeA ETB, Myr ays 2 a= ye AA Gra 2, - [tena ons 260 | 6 Hlu, bye [ Dave Alitrmale +e ond ve Wha Aha\ stent Btu (CG) ty add (Stuck vas 8,5 a,, Of,@ |DaA Stgprmee {an defo Winall Tap | OH, n=O, +H yor N22 [Ass ts H pone ae = Qt) +3 2 [[an-2}* naj oy 2 Oy + Ie 2 (a, +2)*3 - = Oye Ha eH) HH aes ay ee 2Qa,+2t E+ 2 Any 4 (37 + (n-2)h =i) tH Ont At2t3+ SA, tre Bt erent pe hove Oa hor Lt Steere o> 4 ey ; , es ge faa wid es + Oe a ce fotnrte a On ~& . — n(u+) ye ower} + ee % Fiud daw beapbcces Lys of tea stgpstate fined P racwery “4 O=1, On224,_,7! fo ote3 On 1 Oy = we Oy = en") 4 = 24,44 1 OF 2Ay 4 +! =2Poy,ti\+! a (2 tn, +19 +1) +1 = gO Oe! -t oa : PO tay * aS 4 eae] * ee 2 feet 1 + oo we Lowe A HT er f}4 + rp gare eet) ae Lar sand ard Aete - at_ Apt ae 4a" eee fo, mi] amin nl (h-t) WM2 + 2 = oni [4 wy “1i I: 0) 7 n are Kehouaaer Mumbo, & prove thal 4 rs wt AU fositre Aalegurg wv, T=0 Sol (Note? Fikerace Met te ea pad a yO Rot - Frat Fae, fe nze LFrltost Ae ane ey RAF 23 bo ow Let WA AOrdlin } bog 2 Lv | ae - oh jt | Zee hth ~ OWL) = hs By Tha 3 +2 He Vurfies te Aequirsid ActLe got n>) [Baa pt etme se Ata Ie Ba oe, Ay | SR a Fax yy ae wees oo kal _ kg Z Pinks * Bi +f, & | FX Faye t FG | 2h * (Fat Figg, kal ig | 2 fr oF 7 a | ts i YY Fibebae es | Thuy Bs resut dy Lene fee = Ayp Tatas walt os Que bifid — Mouavely 44, ———f 072, 4yef LyaL, cathy fh npr Bins &. Le boshytlys years Lee loth, siterip bg eh, th 2 B+l ey byt bg gee Pipe aag Ly Fhytlgs b +g- 7 lg alittle 224419247 bs clyth o7eh ou 44> Geely 4742957 be = eb eo eT Flog Oy FA ase Rlenases Winn tik Ls Lincas § number, thak bee Fey fr Pa postive Adlpus SL Wedd, ~ fo=9, Fat, FyrGPyta ss ys iach ie . Lo =0, Ly, B SE OET hy eby yo Lye { hel rotls hy ¥ Fe e, biosmslrd= Fy t fF, wreees 3 M's ae Lye alt ok + fy Hue, tr usury Aue for Mel, D2. Arpinme teat tee Curd te Hoe fo Roya, k poe | bi 7 HD he roar, Ly tt tel (hy dy Y Ly) fut) (Sea ¥)ee o baat =( Fic, + Fendt (i, 7 Ef oy gu rs = (fei th fo) 4 (Rt few) Te = = it fia) a a i, a | hes toate abd eat, Pe el a ta Ee al bye faa t faa) . ) 7 [fer te Filenae Sequin, fF Bye cele | eee. oy] 7 se Fe. nz0, he a] bs? f 7 ey | re et ES (33 2) fee | | wie ge Da feat Jr need Assvne taal Abu ett dds Bint for as wre kde A b be fhe emits. 4 Fibovace swies eT) ape Cs (og Ae sptetee)43> ey 2 i. (PTS fest ee 071 sales apt Fog sapere eT | a (> Ene . QF its . atl AE Lee gre) cl EGE ey) tn shaves trae te CF oun pow How, Mh mabmalicas + a Catnanate pow), oud pe i wd now reg ol Ne nkegert we —_——@ Ficyeay - fas fran poste dnfgarie x F se Pres J- kee —a) get 2 eee 9 Doe be pao Leal ye “yl SF \- Fees ee toi 2" me +) A FE. +) Fy. re fia. & Fle ay ; ar gel 2 2 \ CC! ‘ A. #F be NO + : BA VVye dour hal, Faas > Fase t feay _ Recent tus, ke ge & od ” 5 fy Eat = I (Tees) e a : The butt a Sen neem4s> mabtematray MobudBon , wt to proved thay Hema, by n >a Fr) cl- 1 mPa aid —_ — — + —| UNITIS Fun pamper AL PtvepLes of COUNTING @ The Ave of Stet an gia pur Late dow be pufdimed 2. mi! 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She gud 9% fe? a Sol sor Los 4 shia 44 bs v na ae 5 FY rach tb, Cy ts met & Posel btw p eet gud \Crhe ate a pio), (P42, Cb, 9, HP ay es Gna), (asp, CH, CUA, (8) ve (xa), Caddo ¢a,c), 4,4) C4.) i) gggcie an day et od > adn be vane eps a pews Le 00 2 pypnur four ayacbats ies, best re tn To Pt ca digit pd Uo ube gud figt COA vender! Swe f Bagh fettins Wh ue weg cule tale At wo (oyContr 4 “4 4 Spnibols Bee tae eancly Wa TD seledeeg he 0 Ueber Among 74 Meus 4 : Db, kiong ayn guished, Ben ih taarirgt Be uted * pee Aah Lelie tor be Mikal 2. , oq | eg > Fist diyt door OA Stlechid ten IQ ped Se scandy low GES Pe aa hematiinch Bh Styoteat ae Sg [ 26 ¥ 25 XB XI = 58500 naugs: ea pla sustaurout Sells 6 Sorte eee. ee ances | Sadiaen dae, 8 os ea and 2 told beverages, le Der tot anys, ta heen toa Mlk gue Souk dang ane Lk Betray a FL motes Saline ce Pear 0 foth fo [Soke Rust clei te brs che Way CPradusct Luly Suourd theta An gyn 2 9 Lap Chtoduuct, Pole jn. Ae tata vane f hay ba sl cour whet dus [re eh IP EP = 98 (sume Pale) Trae Bus Lords bikin hanes A amd Bane | ue Tee tc Getsern. flows ond ¢. Pa | eabon gf a pie Af fr wo fois las rae sect ot Bee Ovite + Sa At B me 2 pat bE ¢ mes may cash : Coola 2 Lay (blues udt take tee sain Ke A d& (Chaos aay. ea+ he numb J Ways Dra ton pole Me Sours tape ws 4Y¥ 3X9x3 572+ P a Sy at Ya Te Low he Nn ayo 9 ea ws fee 3 “OT vw ey ee Dros oe oy oes ze i She aged Ae, WE ew TE play pit place so git plas wg nef VA Naty as wep x ON nv sre x Ja doi] 4 que tte re Se the 2 Planes fe [he tress are go fous seqptrrees | © Fiad tw nambr f Grunig Styrene oY Leas W sao Brwarg Supine — Ute 0 AK! 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The gen ole Dos 10 Leer nike BA, BSA, od om, Uw a OOS om pa rapa ef pmo OM 10! er = 25) 200 oe spre, * 7 28= ieeeuea |e TY ho Pa as ou Als ane to he one 3 Y NSO ore singe Lew: Rites oe fe & pomudes tus Gag, §, 55 M,U, 4. of pirwilat or dy ce = 940 at ail Fot Fa prgnlatons fo bys wrth og fee kk vn dicrded, the. fou ata 4 prot os & fe, bab tee due Axe 3, four ane 4. “seit HIT — Te 5 How net poste tiliging Lom Lie fre neg Die byt 3,4 ;,6,5,5,67 Fe a 5,000, 000 7 SAN Hoe no mum be gf ta feu Siaee S000 d00 Lay 7 1 bgt Ay Come tabs gstear $f ok-y bo erdeeeal S.ove, 00 accthase ae ® Ey araly, Thue ote to Ih Od eo lar Y Wy 76, ee Hy WM Ay Othovagennenrte Wes Ad 4A Wd ey miy owe Ss. Me 7 WRomginesr dt weer 6) See > 860 "Se ee N= Hy My He Hee MH Yat eaeWe en @ Me Batts 1G %, Wg Mug % a y N=6, dap 4's ord two Me q “ alent iy =~ = {9o ty > 7, he Wunder 7 Operas x ! " 40 Woh» mb . : eek wa Ou: 12 sae 10, Nun bers ae gy antiques 4 nn Mtns mee aes1 Com givaTtions COM BUVAT IONS. 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Sed po lal Dots jad We Noe J Ortangumsnts Without Six ate (4 mITIPPT ) mis, 7 et pute 0 Sy Wbndel toon fe placed Belton o UIs 9M pF q Dag Ta hie ht4 omk BR 7 baad os fe play te SA otis Aa dant ing ern feer ut, a (3 * ap) 4) SEE eee ~ otal rawbir of suck artangunrttt At 165 x “Jo 7 Bso wap Cd284 Graph Theory and Combinatorics 20. Prove the following identities: (i) CQn,2) = 2C(n, 2) +? (ii) CBn,3) = 3C(n,3) + 6nC(n, 2) + (iii) CQnt r+ Ayr) = Cat 1, 1) + Cnt 2,2) + + Cl 7,2) (iv) Cla Lyre 1) = Cin) + Clr + 1,2) Cla) (v) nC(m + n,m) = (m+ 1)C(m+n,m+ |) (vi) CQn+ n,n) = D2 9 Clm,r)C(n,r) (il) Cntrt Lr = Diy Cnt hb Answers 1, C(9,5)x C(15,4) 2. C(20,3) x C(30,4) 3. C(5,4) + C(6.4) 4. C(10, 6) 5. (a) C(5,3) x C(21, 5) x 8! (b) C(4, 2) x C(20, 4) x 6! (c) C(4, 2) x CY. 3) x BE 6. C(21,3)x C(5,2)x 5! 7.450 8 rh 8 ) ‘ZI r+ 9. C(15,3)x3'? 10. 39600 11. 980 12, 7350 13.() 120 (i) 50° Gti) 110 14, (i) 28 (ii) 70. (iii) 2B (iv) 3715. (a) 350 (b) 150 () 105 16. (a) 350 (b)21 (©) 980 (4) 1176 17.32, (a)S (b)10 18.55 19. 16 5.3.1 Binomial and Multinomial Theorems One of the basic properties of C(n, r) = (" is that it is the coefficient of x’y° in r the expansion of the expression (x + y)”, where x and y are any real numbers. In other words, fa (x+y) = 2 (")x» This result is known as the Binomial Theorem for a positive integral index. The numbers (") for r = 0, 1,2,...n in the above result are known as the binomial coefficients. The student is already familiar with the proof by mathematical induction of, the above-mentioned binomial theorem.5. Principles of Counting - I 285 The following is a generalization of the binomial theorem, known as the Multinomial Theorem’. Theorem : For positive integers n and t, the coefficient of x" x'2.x2...x)" in the expansion of (x +X. +--+ + x," is where each n; is a nonnegative integer The Binomial theorem for a positive integral index m reads (x+y) = 2, (iv ie When x = y = I, this becomes Stem S0-tht- = 0 ma and when x = —] and y = |, we get = SC -f)-G) been) : Example 2. Find the coefficient of x°y’ in the expansion of (2x ~ 3y)* » We have, by the binomial theorem, 2 (Qx-3))= y (?) -Qxy-3y)2F a 2 21 : - >| Percale raul In this expansion, the coefficient of x°y* (which corresponds to r = 9) is Io. 99 = 9950 48 yc Bat 99 ye gh y LEX UX 10 (1) Pear = 293? x SAE = a? xa? PO = -2!9x 33 x 11 x 10 = 1946. . 1 Example 3, Evaluate 2 . 2 3,2, 2,5. Principles of Counting - I 287 >» We have 2 12! (323) = Spararay = 166320. 3 Example 4. Find the term which contains x!! and y* in the expansion of (223 — 3x? +296. > By the multinomial theorem, the general term in the given expansion is Jecapstiyne 71, M2, M3, (n.f gfoemeavrner =| Thus, for the term containing x'' and y* we should have 3m +m = 11 and 2nz = 4, so that m = 3 and m = 2. Since m, +n) + n3 = 6, we should then have ny = 1. Accordingly, the term containing x'' and y* is (, 5 Pearse = = (ar ml i x8X 3} xllytz? = = 4320x!!y427_ . 13,2, Example 5. Determine the coefficient of (i) xy in the expansion of (2x -y =2)', and (ii) 632d in the expansion of (a + 2b - 3c + 2d + 5)'®. > (i) By the multinomial theorem, we note that the general term in the expansion of (2x -y—z)* is ( Jeane yy"{-2)" m1, 3, For m = 1, m = Land n3 = 2, this reads (4 / jon Wa = ——— 97 = - 7 = 2x Tee = we. Thus, the required coefficient is -24. (ii) By the multinomial theorem, we note that the general term in the expansion of (a+ 2b - 3c + 2d + 5)'Sis 1 ( : Joranr sen aarisyr Pty Ma, M3, Nay Res,288 Graph Theory and Combinatorics For m = 2, m = 3,m = 2,m = Sands = 16-(2+3+2+5) =4, this becomes 16 ere) 209 pS 16 3 2 x 8 x 54 x glared, 2b)(-3cP (2d)'5* = ~3P xB x St x aD? (o.5.0 5a} cP Qa)s' bososs KBX BP XBx Ss xa? 16! eB x3 x 54 2, HBX¥ XS x Say Bod : 16! : aang x Leap tgs 23x XS x Ge red Thus, the required coefficient is 16! x 25x 53 x3 aye Exercises 1. Find the coefficient of (i) 2y' in the expansion of (x + 2y)"? (ii) x5y° in the expansion of (2x - 3y)’ 2. Show that (,,”,,) = (") =(”)- 3. Compute the following: O65.) Cro) Gd (655, 4. Find the coefficient of (i) xyz in the expansion of (x+y +z)’. (ii) xyz”? in the expansion of (x — 2y + 3z7!)* (iii) 224 in the expansion of (x + y +z)’. (av) 23y*z" in the expansion of (2x — 3y + 52). (v) w°x2y2? in the expansion of (2w — x + 3y ~ 22)°. (vi) x2xyx}xé in the expansion of (xy + x2 + x5 + x4 + x5)!5. Principles of Counting - I 289 5. Prove that, if m is a nonnegative integer, ad +x)" +(1-2)"}= (i) ()e+ : «(i nif nis even where k = ge n-lifnisodd. Deduce that (with k defined as above) are (‘)- 2"! forn > 0 KP \lifn=0. Answers 1. (i) 1760 Gi) 6048 3. (i) 210 Gil) 420 (iii) meaningless (why 2) 4, (i) 42 (ii) -216 (ily 35. Civ) -3024000 (v) 161280 (vi) 12600 5.3.2 Combinations with Repetitions Suppose we wish to select, with repetition, a combination of r objects from a set of n distinct objects. The number of such selections is given by” Cintr-1,n= (" ‘ = wz “(ET ect en- tno n-1 In other words, C(n +r - 1,7) = C(r-+ n= 1,1) represents the number of combinations of n distinct objects, taken r at a time, with repetitions allowed. The following are other interpretations of this number: (i) C(n+r=1,r) = C(r+n-1,n= 1) represents the number of ways in which r identical objects can be distributed among n distinct containers. *For proofs, see Example 11, Section 7.1.2 and Example 8, Section 8.4.2.290 Graph Theory and Combinatorics (ii) C(n +r 1.) = Cr +n—1,n— 1) represents the number of nonnegative integer solutions’ of the equation Xt tet yen Example 1. A bag contains coins of seven different denominations, with at least one dozen coins in each denomination. In how many ways can we select a dozen coins from the bag? > The selection consists in choosing with repetitions, r = 12 coins of n = 7 distinct denominations. The number of ways of making this selection is 18! (7 +12 = 1, 12) = C18: 12) = = 18,564. . 12! 6! Example 2. Jn how many ways can we distribute 10 identical marbles among 6 distinct containers? > The required number is C16 + 10~1, 10) = CCS, 10)= FP = 3003 « Example 3, Find the number of nonnegative integer solutions of the equation Xp + Xy + Xx + Xqt Xs = BL > The required number is C(5 + 8 ~ 1,8) = C(12, 8) = 495. s Example 4. Find the number of distinct terms in the expansion of 6 (xy + XQ +g + XG + > Every term in the expansion is of the form (by multinomial theorem) 16 my am gs ch xt x Ls M25 M34 Ray mg} 234 S TA nonnegative integer solution of the equation x, + x2 ++ +x, =r is an n-tuple (x .X24.3,. 0X9), Where x1, x25. %y are nonnegative integers whose sum is r5. Principles of Counting - 1 291 where each nj; is a nonnegative integer, and these m;s sum to 16. Therefore, the number of distinct terms in the expansion is precisely equal to the number of nonnegative integer solutions of the equation nN +My +3 + Ng +ns = 16. This number is C(S + 16 ~ 1, 16) = C(20, 16) = 4845. . Example 5. Find the number of nonnegative integer solutions of the inequality Xp tx txg te +26 < 10. » We have to find the number of nonnegative integer solutions of the equation Xp tX_ $x3+6° + X6= 9 — x7 where 9 — x; < 9 so that x; is a nonnegative integer. Thus, the required number is the number of nonnegative solutions of the equation Xi tQt Ate $x = 9. This number is a CCT +9 ~1,9) = C(15.9) = sre = 5005 . Example 6. Find the number of integer solutions of xy xy + x3 +X4 + Xs = 30 where x; > 2, x) 23, x3 24, x4 > 2, x5 20. > Let us set yy = x) - 2, 2 = -3, 2 = 49-4, yy = Ka —2, 5 = 5. Then visV2, «Ys are all nonnegative integers. When written in terms of y’s, the given equation reads (+2) + 024+ 3)+ 03 +4)+ (v5 +2) 4ys = 30, or Wty tystystys = 19. The number of nonnegative integer solutions of this equation is the required num- ber, and the number is 23! CG + 19 - 1,19) = (23,19) = Toray = 8855 .292 Graph Theory and Combinatories Example 7. /n how many ways can we distribute 12 identical pencils to 5 chil- dren so that every child gets at least \ pencil? > First, we distribute one pencil to each child. Then, there remain 7 pencils to be distributed. The number of ways of distributing these 7 pencils to 5 children is the required number. This number is ut CS +7-1,7)= CLI) = 37 = 330 2 Example 8. A total amount of Rs. 1S00 is to be distributed to 3 poor studenis A, B, C of a class. In how many ways the distribution can be made in multiples of Rs. 100 (i) if everyone of these must get at least Rs. 300? (ii) if A must get at least Rs. 500, and B and C must get at least Rs. 400 each? > Taking Rs. 100 as a unit, there are 15 units for distribution. In case (i), each of the three students must get at least 3 units. Let us first distribute 3 units to each of the 3 students. Then there remain 6 units for dis- tribution. The number of ways of distributing these 6 units to 4, B, C is the required number (in this case). This number is C(3 + 6 — 1,6) = C(8, 6) = 28. In case (ii), 4 must get at least 5 units, B and C must get at least 4 units each. Let us distribute 5 units to 4 and 4 units to each of B and C. Then there remain 2 units for distribution. Accordingly, the number of ways of making the distribution in this case is C(3 + 2 ~ 1,2) = C(4,2) = 6. s Example 9. /n how many ways can we distribute 7 apples and 6 oranges among 4 children sa that each child gets at least 1 apple? & Suppose we first give 1 apple to each child. This exhausts 4 apples. The remaining 3 apples can be distributed among the 4 children in C(4 +3 - 1,3) = C(6, 3) ways. Also, 6 oranges can be distributed among the 4 children in C(4 + 6 — 1,6) = C(9,6) ways. Therefore, by the product rule, the number of ways of distributing the given fruits under the given condition is 6! 9! (6,3) x (9,6) = sray X Bray = 20x 84 = 1680, . Example 10. Find the number of ways of giving 10 identical gift boxes to 6 persons A, B, C, D, E, F insuch a way that the total number of boxes given to A and B together does not exceed 4.J. Principles of Counting - I 293 > Of the 10 boxes, suppose r boxes are given to A and B together. Then 0 A binary number that contains (7 ~ 1) 1’s and r 0°s, has — 1 +r positions and is determined by positions of 0’s, The number of such binary numbers is therefore C(n~ 1 +7,7). a Example 13. Given positive integers m,n with m = n, show that the number of ways to distribute m identical objects into n distinct containers such that each container gets at least r objects, where r < (m/n), is Con-1+U-rnayn- 1). » Suppose we place r of the #7 identical objects into each of the n distinct con- tainers. Then, there remain (sm — nr) identical objects to be distributed into 7 distinct containers. The number of ways of doing this is the required rumber. This number is C(n + (m— nr) = 1,m= nr) =C(n+ (m—nr)-1,2- 1 =C(m-1+(1-nn-) 2 Exercises 1. In how many ways can 20 similar books be placed on S diferent shelves? r . Find the number of ways of placing 8 identical balls in 5 numbered boxes. e . Determine the number of nonnegative integer solutions of the equation x) +x) + 5+ x4 = 7. & . Find the number of distinct terms in the expansion of (w + x+y + 2). 5. How many integer solutions are there to x, + £2 + x3 + xy +.x5 = 20 where each mz 2? ry . How many integer solutions are there to x; + x2 + x3 + x4 + x5 = 20, where x; > 3, Xp > 2,x3 24, x4 > 6, x5 > 0? ~ ‘Recall that C(n.r) = C(n,n—7).