ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 1 - I S T HP NI DUNG Phn 1: Hon v, chnh hp, t hp v ng dngI.L thuyt Qui tc cng, qui tc nhn. Hon v, chnh hp; t hp.II.Cc dng ton ng dng Dng 1: Rt gn biu thc i s t hp.Dng 2: Chng minh ng thc; bt ng thc i s t hp. Dng 3: Gii phng trnh; bt phng trnh i s t hp.Dng 4: Cc bi ton m s phng n.Phn 2: Nh thc Newtn v ng dng I.L thuyt:Nh thc NewtnII.Cc dng ton ng dng Dng 1:Tnh tng t hp. Dng 2:Chng minh ng thc; bt ng thc i s t hp.Dng 3:Xc nh h s ca s hng trong khai trin nh thc Newtn.

Phn I: HON V, CHNH HP, T HP V NG DNG I.L thuyt: Qui tc cng, qui tc nhn. Quy tc cng: Nuc m1 cch chn i tng x1, c m2 cch chn i tng x2,...mn cch chn i tng xn v nu cch chn i tng xi khng trng vi i tng xj no( i khc j; i, j = 1,2,....,n) th c m1 + m2 +....+ mn cch chn mt trong cc i tng cho. Quy tc nhn: Nu mt php chn c thc hin qua n bc lin tip, bc 1 c m1 cch, bc 2 c m2 cch,.... bc n c mn cch, th php chn c thc hin theo m1.m2...mn cch khc nhau. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 2 - II.Hon v, chnh hp; t hp. 1. Hon v: Cho tp hp A gm n phn t ( n_1). Mi cch sp th t n phn t ca tp hp A c gi l mt hon v ca n phn t . S hon v ca n phn t :Pn = n! = 1.2.3.4.5.n ( ; n N \ _ n 1) ; Qui c0! 1 = . 2. Chnh hp: Cho tp hp A gm n phn t. Mi b gm k (1_ k_n) phn t sp th t ca tp hp A c gi l mt chnh hp chp k ca n phn t ca A. S chnh hp chp k ca n phn t :( 1)...( 1)knn n n kA= (1_ k _n) !( )!nkAnn k=

(1_ k _n) 3. T hp: Cho tp hp A gm n phn t. Mi tp con gm k (0_ k_n) phn t ca A c gi l mt t hp chp k ca n phn t cho. S t hp chp k ca n phn t : !!( ) !knnk n kC=(0_ k _n) Cc dng ton thng gp: Dng 1: Rt gn biu thc i s t hp.Dng 2: Chng minh ng thc; bt ng thc i s t hp. Dng 3: Gii phng trnh; bt phng trnh i s t hp.Dng 4: Cc bi ton m s phng n. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 3 - Dng 1: Rt gn biu thc i s t hp.Phng phpS dng cc cng thc sau rt gn biu thc i s t hp: !nP n =( *n N ) !( ) !nkAnn k=(1_ k_n) !!( ) !nkCnk n k=(0_ k_n) Mt s v d V d 1: Rt gn biu thc:1!2nkAnkk=_= Bi gii Ta c nhn xt: ( )1 1 1! ! 1 !kk k k= Suy ra ( )1 1 1 1 1 1 1 1... 1! 1! 2! 2! 3! ! ! 1 !2nkAnk n n nk= = = _= V d 2: Rt gn biu thc:6 54A An nAAn=Bi gii Ta c 2( 1)...( 5) ( 1)...( 4)4 ( 4)( 5) ( 4)( 1)...( 3)n n n n n nA n n n nn n n = = =

V d 3: Rt gn biu thc:212 ...1 1nC Cn nA C nnnC Cn n=

Bi gii: Ta ln lt c: 1nC n =ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 4 - 1!22!.( 2) !2 2.1 !1!.( 1) !nnCnnnCnn= = ...111 !1!.( 1) !nCnnn nCnn= =

( 1)1 ... 2 1 .2n nA n n= = = Bi tp t luyn:Bi 1: Rt gn biu thc: 11( 1)nnkCk k== _

Bi 2: Rt gn biu thc: ( 1) ! 5!( 1) 3!( 1)!mAm m m= . Bi 3: Rt gn biu thc: 10 9 12 1117 17 49 4910 849 17A A A ABA A = - Bi 4: Tnh gi tr biu thc 4 3 4 27 7 8 35 6 62 10 10 1111C C C ASP C C C =

Dng 2: Chng minh ng thc, bt ng thc i s t hp Phng php: Thc hin cc bc sau + S dng cc cng thc: !nP n =( *n N ) !( ) !nkAnn k=(1_ k_n) !!( ) !knnCk n k= (0_ k _n) ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 5 - -1(0 )-1 -1k k kC C C k nnn n= < < a ng thc, BT i s t hp thnh ng thc, bt ng thc i s thng thng. + Chng minh ng thc, bt ng thc i s thng thng suy ra pcm. Mt s v d: V d 1: CMRvi k, n N, 3 k n ta c: 2 1 2 n n nA A kAn k n k n k = Bi gii ( ) ! ( )!2 1( 2) ! ( 1) !( )! 1 ( ) !1( 2)! 1 ( 1)( 2) !2 2( ) ! ( ) !2( 1).( 2) ! !VPn k n kn nVT A An k n kk kn k k n kk k k kk n k k n knkAn kk k k k 1 =

( )= = = = = = = V d 2: Chng minh rng: 12 2( !) ( ) ( , 2)2nn nn n n Z n< < (1) Bi gii Bin i BT (1) v dng: 12 2(1.2.3.... ) ( )2nn nn n< <

21 2[(1. )2.( 1)].3.( 2)..... ( 1)] ( )2nn nn n n n k n k= < < (2) a)Ta c nh gi: ( 1) ( , 1) k n k n k n k _ \ < (* )

Do (* ) ( 1) ( 1) 0 ( )( 1) 0 n k k k n k k = = _ ng , 1 k n k \ < p dng BT (*) vi k = 2,, n -1 ta c ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 6 - 1.2.( 1)....( 1)....1n nn nk n k nn n'111111111!111111111+_ _nbt ng thc. (a)2[(1. )2.( 1)].3.( 2).....( 1)...( 1).2( .1)] nnn n n k n k n n= b) S dng BT Csi tac : 1 12 2( 1) ( ) ( )2 2k n k nk n k _ =(**) 0, k n k \ _ _p dng BT (**) vi k =1,2,, n ta c 222( 1)1212211.2212........( 1)........( 1)2211.2nnnnnnnk n knnn'1 l 11 l11l1 l111 l11l _1 l 11l111111 l 11 l! l11l111111l1 l11 l1 l111 l11 l11l1 l11+_ _ __ BDTSuy ra 21(1. )[2.( 1)].3.( 2).....( 1)...( 1)2( .1)2nnn n n k n k n n l l l l < (b) ( ) () a b =pcm ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 7 - V d 4: CMR a) 1 2 33 33k k k k kC C C C Cn n n nn = b) -1 -22 (2 )2k k k kC C C C k nn n nn= _ _ c) 1 2 3 3 22 5 43 2k k k k k kC C C C C Cn n n nn n = Bi gii a) Ta c) )1 1 2 2 3( 2( ) (1 2 1 1 22 ( ) ( )1 1 1 1 1 1 112 2 3k k k k k kVT C C C C C Cn n n n n nk k k k k k kC C C C C C Cn n n n n n nk k kC C C VPn n n = = = = = = b) Ta c:-1 -22 (2 )2k k k kC C C C k nn n nn= _ _ Nn

- 1 - 1 - 2 - 11 12k k k k k kVP C C C C C Cn n n n n nkC VTn 1 1 1

( ) ( ) ( ) 1

( )= = = = c) 1 1 2 3) )12 3( (k k k kn n n nC C C Ck kVT C Cn n 1

( )=

1 2 32 31 1 1k k kC C Cn n n 1 1 1

( ) ( ) ( ) =

1 2 2 321 1 1 12 322 22 2 3 2 32 2 3 2 2k k k kC C C Cn n n nk kC Cn nk k k k kC C C C C VPn n n n n 1 1 ( ) ( ) 1 1 ( ) ( ) 1 1 1

( ) ( ) ( ) = = = = = ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 8 - V d 5: CMR:2. ( ) (0 ) (1)2 2 2-n n nC C C k nn n k n k_ _ _ Bi gii: Ta c ( )( )( )( )( ) ( 2. ( ) 0 ) (1)2 2 2-22 ! 2 ! 2 !.!. ! !. ! !. !( 1)...( ) ( 1)...( )n n nC C C k nn n k n kn k n k nn n n k n n k nn k n k n n k n k n 1

( ) ll= ll ll_ _ _ = _

2( 1)...( )( 1)( 2)...( ) ( 1)( 2)...( )( 1)...(n n nn k n k n k n n k n k n k nn n l l l ll= ll ll_ _

...2)( 1)( 1) ( )( )2( 1)...( ) (*)nn k n k n k n n k nn n n l l l ll ll ll l l l= _ TheoBT Cauchy ta c( )2( )( ) n k i n k i n i \ _ _ _ 0 k n;i = 1...nCho 1, i n = ta c BT (*) Vy BT (*) ng (1) c chng minh. Bi tp tng t Bi 1: 1. ( 1) ( , )n mnC m C m n Zmn mn =

Bi 2:2 2( 2, )1C C n n n Znn= _ Bi 3: -. . (, , , , )-r k k r kC C C C r k n Nr n k rn r n n r= _ _ Bi 4: 12 -1 1( )2 2 2 22n nC C C n Zn n n = Bi 5: -1-1nr rC Cnnr= Bi 6*: ( )2 3112. 3 ... ....1 2 - 1 - 1 2p nC C C Cnnn n n nC p nnp nC C C Cn n n n =ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 9 - Bi 7: , 2 n Nn \ _ ta c ( )11 1 1...2 2 22 3nnA A An = Bi 8: CMR:-2( - 1) ( 1)-2k kk k C n n Cnn= Bi 9: CMR:-1(0 )-1 -1k k kC C C k nnn n= < < Bi 10: CMR: 2. ( ) 0 )(1)2 2 2-n n nC C C k nn n k n k_ _ _

Bi 11: CMR:11 1k k kA A kAnn n= Bi 12: CMR:1 2 3 12 3 ... 1n nP P P nP P = Bi 13: CMR: 1 2 3 2 32 5 42 3k k k k k kC C C C C Cn n n nm m = Bi 14: CMR: 1 22 (2 , )2k k k kC C C C k nn n nn = _ _

Bi 15: CMR: . . . ( , ; , , )r k k n kaC C C C r n k r n r k Zn r nr k= _ _ 1. ( )11 2 1. ( )1 2 1nrr rbC C r nnrr r r rcC C C C r nnn n r==< < + +...+ Bi 16: CMR:

2) 3) 4) 0 1 1 2 2 5 51) . . . ... .5 5 5 5 51 2 1....1 2 11 2 3 44. 6. 4.43 1 23. 33k k k k kC C CC C C C C Cn n n nnr r r rC C C Cnn n rk k k k k kC C C C C Cn n n n nnk k k k kC C C C Cn n n nn = = = =

1 1215)2m m mn n nmC C C Cn =

3 1 23 331 2 3 2 35. 4.2 3k k k k kC C C C Cn n n nnk k k k k kC C C C C Cn n n nn n == 6)7)2 Bi 17: CMR: 10002001 2001 2001 20011 1001 k kC C C C _ _ _ (0 k 2000)( HQGHN A 99- 00 ) Bi 18: CMR: 100 1002 250102 102100C < a. Ta c: 5 cch chn ng i t X n Y, ng vi mi cch chn ng c 4 cch chn ng i t Y n Z.Do , c tt c: 5. 4 = 20 cch chn ng i t X n Z qua Y b. Theo a) c 20 cch chn ng i t X n Z qua Y Khi tr v ng vi mi cch chn ng , t Z n Y c3 con ng chn, do c 3 cch. ng vi mi cch chn ng , t Y v X ch cn li 4 cch chn.Do , c tt c 3. 4 = 12 cch chn ng i v t Z n X qua Y. Vy c tt c: 20. 12 = 240 cch chn ng i v trn tuynX Z qua thnh ph Y bng nhng con ng khc nhau < 22 > Ta thy: - c 35 cch chn trng i hc - C 25 cch chn trng cao ng - C 21 cch chn trng trung hc chuyn nghip Khi chn thi trng i hc th khng chn trng thi l cao ng v chuyn nghip, tng t vi cao ng v trung hc chuyn nghip, do c tt c: 35 + 25 + 21 = 81 cch chn trng thi. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 20 - Nhn xt: Nhiu bi ton m phc tp khng th gii c nu ch s dng hoc quy tc nhn hoc quy tc cng. Nhng chng ta c th gii c nu s dng c 2 quy tc ny.< 23 > Gi P l tng s mt khu c th v 6 7 8, , PPPtng ng l s mt khu di 6, 7, 8 k t. Theo quy tc cng ta c: 6 7 8P P P P = Ta s tnh6 7 8, , PPP: - Tnh 6P : S xu di 6 k t l ch in hoa hoc ch s l:636 . V mi v tr c 36 cch chn. S xu di 6 k t l ch in hoa v khng cha ch s no l:626 .Vy: 6 6636 26 P= - Tng t: 7 7736 26 P= 8 8836 26 P= Vy, ta c: 6 7 8P P P P = = 2684483063360 < 24 > C 4 loi sch, do c 4! Cch sp xp theo mn. mi loi sch c: 3! Cch sp xp sch ton. 4! Cch sp xp sch l 2! Cch sp xp sch ho 5! Cch sp xp sch sinh Vy c tt c: 4!. 3!. 4!. 2!. 5! = 829440 cch sp xp. < 25 > Mi cch chn bn cu th ca i bng l chnh hp chp 4 ca 10 phn t.Ta c: 4105040 A =cch chn < 26 >S cch tng sch l s cch chn 6 cun sch t 9 cun c k th t. Vy s cch tng l: 6960480 A =1.Ta nhn xt rng: khng th chn sao cho cng ht 2 loi sch. + S cch chn 6 sch t 12 sch l: 612665280 A =+ S cch chn sao cho khng cn sch vn: 5 16 7. 5040 AA =+ S cch chn sao cho khng cn sch nhc: 4 26 8. 20160 AA =+ S cch chn sao cho khng cn sch hi ho: 3 36 9. 60480 AA =ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 21 - + S cch chn cn tm l: 665280 85680 = 579600 < 27 > Ban cn s lp gm 3 ngi trong lp khng c s sp xp a.Mi mtbancns 3 ngil mttpcon 3 phntca tp hp 40 hcsinhca lp. Vy c: 3409880 C =cch lp ban cn s lp 3 ngi. b.C 125Ccch chn 1 hc sinh nam v 215Ccch chn 2 hc sinh nam.Do c 1 225 15. 2625 C C =cch lp mt ban cn s lp gm 1 nam v 2 n c.C 315455 C =cch chn 3 n sinh nn c 455 cch lp ban cn s lp 3 ngi ton n.D c: 9880 455 = 9425 cch lp ban cn s 3 ngi ma trong c t nht mt nam < 28 > Nhn xt: Vic phn cng vo 1 tnh khng c s sp xp C 1 43 12. CCcch phn cng cc thanh nin tnh nguyn v tnh th nht. Vi mi cch phncngccthanhnintnhnguynvtnhthnhtthc 1 42 8. CC cchphncngcc thanh nin tnh nguyn v tnh th 2. Vi mi cch phn cng cc thanh nin tnh nguyn v tnh th 1 v tnh th 2 th c 1 41 4. CC cch phn cng cc thanh nin tnh nguyn v tnh th 3.S cch phn cng cc thanh nin tnh nguyn v 3 tnh tho mn yu cu bi ton l: 1 4 1 4 1 43 12 2 8 1 4. . . . . 207900 CC CCCC =< 29 > Bn Hoa c 2 cch chn bng cm bnh nh sau: Cch 1: Chn 2 bng hng bch v 3 bng hng nhung + S cch chn 2 bng hng bch trong 10 bng: 210C+Vimicchchn2bnghngbchlic 310C cchchn3bnghngnhung trong 10 bng.Vy cch 1 c210C . 310C cch chn bng.Cch 2: Chn 3 bng hng bch v 2 bng hng nhung. Lp lun tng t nh trn, ta cng c 210C . 310Ccch chn bng.Vy bn Hoa c s cch chn bng l:3 210 102 10800 CC =cch chn < 30 > Nhn xt: Ni dung khng ph thuc vo vic sp xp th t cu hi. Mi kim tra phi c s cu d l 2 hoc 3, nn c cc trng hp sau: ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 22 - - c 2 cu d, 2 cu trung bnh, 1 cu kh, th s cch chn l: 2 2 115 10 5. . 23625 C C C = - c 2 cu d, 1 cu trung bnh, 2 cu kh, th s cch chn l:2 1 215 10 5. . 10500 C C C = - c 3 cu d, 1 cu trung bnh, 1 cu kh, th s cch chn l: 3 1 115 10 5. . 22750 C C C =V cc cch chn trn i mt khc nhau, nn s kim tra c th lp c l: 23625 + 10500 + 22750 = 56875 B-Bi ton m s phng n c lin quan n hnh hc. Bi 1: Cho 7 im trn mt phng sao cho khng c 3 im no thng hng a. C bao nhiu ng thng m mi ng thng i qua 2 trong 7 im ni trn? b. C bao nhiu tam gic vi cc nh l 3 trong 7 im ni trn ? Bi 2: Tm s giao im ti a ca : a.10 ng thng phn bit? b. 6 ng trn phn bit? c.10 ng thng v 6 ng trn trn? Bi 3:a. C bao nhiu ng cho trong mt a gic li n cnh? b. C bao nhiu tam gic c nh l nh ca a gic n cnh? Trong c bao nhiu tam gic c cnh khng phi l cnh ca a gic n cnh ? Bi 4: (H, C Khi B 2003) Cho a gic u 1 2 2... ( 2, )nAA A n n Z _ ni tip ng trn (O). Bit rng s tam gic c nhl 3trong2nim 1 2 2, ,...,nAA A nhiugp20lnshnhchnhtcccnhl4 trong 2n im 1 2 2, ,...,nAA A , tm n.Bi5:(HCSND-1999-2000):ChotamgicABC,xttphp4ngthngsong songviAB,5ngthngsongsongvi BCv6ngthng songsongviAC.Hi cc ng thng ny to c: a. Bao nhiu tam gic ? b. Bao nhiu hnh thang ( Khng k hnh bnh hnh ) ? c. Bao nhiu hnh bnh hnh ? Bi 6: ( CSP -A- d b - 02 02 ): Cho a gic li n cnh. Xc nh n a gic c s ng cho gp i s cnh ? ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 23 - Bi 7:(HNT- 01 02):Trong mt phngchothp gicli 1 2 10... AA A .Xtttccc tam gic m 3 nh ca n l nh ca thp gic. Hi trong s cc tam gic c bao nhiu tam gic m 3 cnh ca n u khng phi l cnh ca thp gic ? Bi 8: ( HVNH- D- 2000 ):Trong mt phng cho a gic u (H) c 20 cnh. Xt cc tam gic m 3 nh ca n ly t cc nh ca (H). 1. C tt c bao nhiu tam gic nh vy? C bao nhiu tam gic c ng 2 cnh l cnh ca(H) ?2. C bao nhiu tam gic c ng 1 cnh l cnh ca(H) ?3. C bao nhiu tam gic khng c cnh no l cnh ca(H) ? Bi 9: Cho 2 ng thng song song.Trn ng th nht c 10 im. Trn ng th hai c 20 im. C bao nhiu tam gic to bi cc im cho ? Bi10:(HC-B-2002):Choagicu 1 2 2...nAA A (2 n _ ,nnguyn)nitip ng trn ( 0). Bit rng s tam gic c cc nh l 3 trong 2n im1 2 2, ,...,nAA Anhiu gp 20 ln s hnh ch nht c cc nh l 4 trong 2n im 1 2 2, ,...,nAA A , tm n. Li gii: < 1 >a. Mi cp im khng k th t, trong 7 im cho xc nh mt ng thng v ngc li. Vy, s ng thng i qua 2 trong 7 im ni trn bng:

( )277! 6.7211.2 2! 7 2!c = = = ng thng. b.Mi b 3 imkhngktht,trong 7 im choxc nh mttamgicv ngc li. Vy s tam gic c nh l 3 trong 7 im ni trn bng: < 2 >a. Hai ng thng phn bit c ti a 1 giao im. S giao im ti a ca 10 ng thng phn bit l s t hp chp 2 ca 10, do bng: ( )21010!452!10 2!C = = im b. Hai ng trn phn bit c ti a 2 giao im. S giao im ti a ca 6 ng trn phn bit gp 2 ln s t hp chp 2 ca 6, do bng: 262. 2.15 30 C = =im.ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 24 - c. Mt ng thng ct mt ng trn ti a ti 2 im. Do s giao im ti a ca 10 ng thng phn bit vi 6 ng trn phn bit bng: 10.6.2=120 im Khi , s cc giao im bng: 45 + 30 +120 = 195 im < 3 >a. Ta c:* Mi a gic li n cnh th c n nh.*Mionthngni2nh btk,khngktht,thhocl1cnh, hoc l mt ng cho ca a gic .Vy s ng cho ca a gic n cnh bng: 2nC n b. S tam gic c nh l nh ca a gic n cnh l s t hp n chp 3: ( )3! ( 1)( 2)6 3! 3!nn n n nCn = = S tam gic c t nht 1 cnh l cnh ca a gic gm 2 loi: * S tam gic ch c 1 cnh bng 14 nC

* S tam gic 2 cnh bng 1nCSuyra,stamgicc1hoc2cnhcaagicl: 1 1 1 1 1 14 3 3. ( 4) ( 3) . .n n n n n nC CC n n n n n nC CC = = = =Vy, s tam gic c cnh khng phi l a gic l:

23 1 13( 2)( 1) ( 2)( 1) 6( 3) ( 9 20). ( 3)6 6 6n n nn n n n n n n n n n nC CC n n = = = < 4 > S tam gic c cc nh l 3 trong 2n m 1 2 2, ,...,nAA Al 32nCGi ng cho ca a gic u 1 2 2... ( 2, )nAA A n n Z _ i qua tm ng trn (O) l ng cho ln th a gic cho c n ng cho ln.Mi hnh ch nht c cc nh l 4 trong 2n im 1 2 2, ,...,nAA Ac cc ng cho l 2 ng cho ln. Ngc li, vi mi cp ng cho ln ta c cc u mt ca chng l 4 nh ca 1 hnh ch nht. Vy s hnh ch nht ni trn bng s cp ng cho ln ca a gic 1 2 2...nAA A , tc 2nC .Theo gi thit th:ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 25 - ( ) ( )3 22(2) ! ! 2(2 1)(2 2) ( 1)20 20 206 2 3! 2 3 ! 2! 2!n nn n n n n n nC Cn n = = = = = = 2 1 15 8 n n = = = = .

C-Bi ton m s phng n lin quan n s t nhin Ghi nh: Khi gii bi ton m lin quan n s t nhin ta cn lu : Nm vng qui tc cng nhn. Ta thng gi s t nhin cn tm l 1 2 3...nn aa a a =sau cn c vo u bi i chn tng ch s mt.S no yu cu cao th chn trc. Khi chncc ch s cn phn tch cu vn u bi cn k, nm chc bn cht ca tng i tng t tm tp xc nh v cc kh nng c th. Cn thn khi c s 0. Phi lun lun ngh ti phn b, nu phn b n gin hn ta tm phn b trc. Bi tp: Bi 1: ( HQG HCM - 99) Vi cc s 1,2,5,7,8 c th lp c bao nhiu s gm 3 ch s phn bit tho mn iu kin: a. L 1 s chn. b. L 1 s nhhn hoc bng 278. c. L 1 s chn v nhhn hoc bng 278. Bi 2: Xt mt dy s gm 7 ch s ( Mi ch s c chn t cc s 0, 1 ,2, 3, 4,9 ) tho mn tnh cht: - Ch s v tr th 3 chn. - Ch s v tr cui cng chia ht cho 5. - Cc ch s v tr th 4, 5, 6 i mt khc nhau. Hi c tt c bao nhiu dy s nh vy ? Bi 3: ( HCSND 99- 99):vi 10 ch s t 1-->9 c th lp c thnh bao nhiu ch s gm 5 chs khc nhau? Bi 4: (HSPV28-99-00): Cho 8 ch s 0, 1, 2, 3, 4, 5, 6, 7. T 8 ch s trn c th lp c bao nhiu s, mi s gm 4 ch s, i mt khc nhau v chia ht cho 10. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 26 - Bi 5: (HSPV - B - 99- 00): C 5 s t nhin 1, 2, 3, 4, 5 . Hi c bao nhiu s c 3 ch s i mt khc nhau to thnh t 5 s cho? Bi 6: ( HYHN 99 00) C th lp c bao nhiu s chn c nm ch s khc nhau ly t cc s 0, 2, 3, 6, 9, Bi 7: ( CSPHN - 36)C 5 ming ba, trn mi ming c ghi 1 trong 5 ch s 0, 1, 2, 3, 4. Ly 3 ming t 5 ming ba ny t ln lt cnh nhau t tri qua phi c cc s gm 3 ch s. Hi lp -c bao nhiu s c ngi gm 3 ch s v trong c bao nhiu s chn. Ch : cc ch s i mt khc nhau do mi s ch ch c mt ming ba. Bi 8: (HQGHCM - 3 00 01) 1. C bao nhiu s chn gm 6 ch s khc nhau i mt trong ch s u tin l ch s l? 2. C bao nhiu s gm 6 ch s khc nhau i mt trong c ng 3 ch s l v 3 ch s chn. Bi 9: HSPHN2 - 8): C th lp c bao nhiu s gm 8 ch s t ctrong cc ch s 1 v 6 u c mt 2 ln cc ch s khc c mt mt ln. Bi 10: Vi cc s 0, 1, 2, 3, 4, ,5 ta c th lp c bao nhiu s gm 8 ch s trong s 1 c mt 3 ln mi s khc c mt 1 ln Bi 11: ( HHu 00 01 - 26) Cho cc ch s 0, 1, 2, 3, 4, 5 t cc ch s cho lp c: 1. Bao nhiu ch s chn c 4 ch s v 4 ch s khc nhau i mt? 2. Bao nhiu ch s chia ht cho 5, c 3 ch s v 3 ch s khc nhau i mt. 3. Bao nhiu ch s chia ht cho 9, c 3 ch s v 3 ch s khc nhau i mt. Bi 12: Ngi ta vit cc s c 6 ch s bng cc ch s 1, 2, 3, 4, 5 nh sau: Trong mi s c vit c mt ch s xut hin 2 ln cn cc ch s cn li xut hin 1 ln. Hi c bao nhiu s nh vy. Bi 13: C bao nhiu s t nhin c 7 ch s c vit bi duy nht 3 ch s 1, 2, 3, trong ch s 2 xut hin 2 ln. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 27 - T luyn Bi 14: C bao nhiu s t nhin c 3 ch s ? Bi 15:ChoA={1,3,5,6,8}.C bao nhiust nhingm 4chslytccchs trong tp A ? Bi 16: Cho A = {0,1,2,3,5,7,9}.T tp A c bao nhiu s t nhin gm 4 ch s i 1 khc nhau. Bi 17: Vi tp E = {1,2,3,4,5,6,7} c th lp c bao nhiu s gm 5 ch s phn bit v: a. Trong c ch s 7. b. Trong c ch s 7 v ch s hng ngn lun l ch s 1. Bi 18: Vi 5 ch s 1,2,3,4,5 c th lp c bao nhiu sgm 5 ch s i 1 khc nhau: a. Khng bt u t ch s 1 b. Khng bt u t 123. Bi19:ChoE={0,1,2,3,4,5,6,7}.Cthlpcbao nhiusgm5chsi1 khc nhau ly t E trong mi trng hp sau: a. L s chn. b. Mt trong 3 s u tin bng 1. Bi 20: T cc s 0,1,2,...,9 c th lp c bao nhiu sgm 6 ch s khc nhau sao Cho trong cc s c mt ch s 0 v 1. Bi 21:Viccchs 0,1,2,3,4,5tacthlpc bao nhiusgm 5ch skhc nhau sao cho trong cc s phi c mt ch s 5. Bi 22: Cho cc ch s 0,2,4,5,6,8,9. Hi c th lp c bao nhiu s: a. C 3 ch s m trong mi s cc ch s khc nhau. b. C 4 ch s khc nhau v c ch s 5. Bi 23: Vi 5 ch s 1,2,3,4,5 c th lp c bao nhiu s gm 5 ch s phn bit v tho mn: a. Mi s nh hn 40000. b. Mi s nh hn 45000. Bi24:Choccs0,1,2,...,9cbaonhiuslgm6chskhcnhaunhhn 60000 xy dng t 10 ch s . ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 28 - Bi 25: T cc s 0,1,2,3,4,5,6 c th lp c : a. Bao nhiu s t nhingm 5 ch s khc nhau. b. Bao nhiu s t nhinchn gm 5 ch s khc nhau. Bi 26: C bao nhiu s t nhin gm 3 ch s khc nhau c th lp c t cc ch s 0,2,4,6,8. Bi 27: T cc s 0,1,2,3,4,5,6 c th lp c : a. Bao nhiu s t nhingm 5 ch s 6,7.b.Tmcc s t nhin gm 5 ch s ly t 7 s trn sao cho: 1. Ch s u tin l 3. 2. Cc ch s u khc nhau. 3. Khng tn cng bng ch s 4. Bi 28:Viccchs 0,1,2,3,4,5tacththnhlp c bao nhiuschn mis gm 5 ch s khc nhau. Bi 29: Cho tp A ={0,1,2,3,4,5,6,7,8}. T tp A: a. C th lp c bao nhiu s t nhin l gm 3 ch s. b. C th lp c bao nhiu s t nhin gm 3 ch s l. c. C th lp c bao nhiu s t nhin chn gm 4 ch s i 1 khc nhau. d. C bao nhiu s t nhin gm 5 ch s i 1 khc nhau v chia ht cho 5 e. C th lp c bao nhiu s t nhin gm 6 ch s i 1 khc nhau sao cho ch s ng cui chia ht cho 4. Bi 30: Vi 4 ch s 1,2,3,4 c th lp c bao nhiu sc cc ch s phn bit. Bi 31: Vi 5 ch s 1,2,3,4 ,5 c th lp c bao nhiu s gm 5 ch s phn bit v l a. S l. b. S chn. Bi 32: (HAN - 97 )T 7 ch s 0,1,2,3,4,5,6 c th lp c bao nhiu s chn c 5 ch s khc nhau. Bi 33: T 6 ch s 0,1,2,3,4,5 c th lp c bao nhiu sgm 4 ch s khc nhau trong c bao nhiu s chia ht cho 5. Bi34:T8chs0,1,2,3,4,5,7,8cthlpcbaonhiusgm4chskhc nhau v khng chia ht cho 5. ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 29 - Li gii Cch 1: t E = {1,2,5,7,8 }. Gi s t nhin gm 3 ch s ln=1 2 3a a a (10 a= ) a. Do nchn nn a3 {2,8}= a3 c 2 cch chn a1 E \ {a3}= a1 c 4 cch chn a2 E \ {a1,a3}= a2 c 3 cch chn Vy: c 2.3.4 = 24 cch chn hay c 24 s. Cch 2: a. Donchn nn a3 {2,8}=a3 c 2 cch chn a1, a2 l 1 b phn bit th t c chen t E\{a3} do n l mt chnh hp chp 2 =24Acch chn. Theo qui tc nhn, s cc s chn gm 3 ch s phn bit hnh thnh t tp E bng 2. 24A= 24 (s). b. Donnh hn 278 nn a1 {1;2}. Trng hp 1: Nu a1 = 1 th a2 E\{a1}= a2 c 4 cch chn a3 E \ {a1,a2}= a3 c 3 cch chn = c 1.4.3 = 12 cch chn . Trng hp 2: nu a1 = 2 th a2 E\{2,8}= a2 c 3 cch chn a3 E \ {a1,a2}= a3c 3 cch chn = c 1.3.3 = 9 cch chn . Vy: c 12 + 9 = 21 cch chn s c 3 ch s phn bit v nh hn 278. Tc l c 21 s tho mn ycbt. c. Donchn nn a3 {2,8} v s cn tm nh hn 278 nn a1_ 2. Trng hp 1: nu a1 = 2= a1 c 1 cch chn a3{2,8}=a3 c 2 cch chn a2 E \ {a1,a3}= a2 c 3 cch chn = c 1.2.3 = 6 cch chn . Trng hp 2: nu a1 = 2= a1 c 1 cch chn a3{2,8}\{a1}=a3 c 1 cch chn a2 E \ {a1,a3}= a2 c 3 cch chn = c 1.1.3 = 3 cch chn . ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 30 - Vy: c 6 + 3 = 9 cch chn s t nhin chn gm cc ch s khc nhau v nh hn hoc bng 278. Tc l c 9 s tho mn ycbt. < 15 > Cc s t nhin l: 0,1,2,3,4,5,6,7,8,9. Gi s t nhin c 3 ch s cn tm l : 1 2 3a a a (10 a = ) Do a1 {1,2,3,4,5,6,7,8,9} nn a1 c 9 cch chn. Do a2 {0,1,2,3,4,5,6,7,8,9} nn a2 c 10 cch chn. Do a3 {0,1,2,3,4,5,6,7,8,9} nn a3 c 10 cch chn. Vy: c 9.10.10 = 900 cch chn hay c 900 s tho mn ycbt. < 16 > Gi s t nhin c 4 ch s cn tm l 1 2 3 4a a a a (10 a= ) Ta thy: a1 c 5 cch chn. a2 c 5 cch chn. a3 c 5 cch chn. a4 c 5 cch chn. Do theo qui tc nhn c: 5.5.5.5 = 625 cch chn hay c 625 s tho mn. < 17 > Gi s t nhin gm 4 ch s i 1 khc nhau l: 1 2 3 4a a a a (10 a= ) Do a1 khc 0 nn a1 c 6 cch chn. Sau khi chn a1 cn 6 s t nhin nn a2 c 6 cch chn. Tng t, a3 c 5 cch chn. a4 c 4 cch chn. Vy:c 6.6.5.4 = 720 cch chn hay c 720 s tho mn ycbt. Gi s cn tm l 1 2 3 4 5a a a a a (10 a= ) Gi s a1 = 7 khi s cn tm c dng: 2 3 4 57a a a a.... V a2, a3, a4, a5 l 1 b phn phn bit th t c chn t E\{7}nn c 466!36021A = =

s. Do s 7 v tr bt k nn ta c th i ch ca cc v tr a1, a2, a3, a4, a5 . Vy c 465.A = 1800 cch chn hay c 1800 s tho mn ycbt. b. Gi s cn tm l 1 2 3 4 5a a a a a (10 a= ) Do ch s hng nghn bng 1 nn a2 = 1 ST: L Ngc Sn_SP Ton K07_H Ty NguynChuyn : iSTHp - 31 - Chn 1 v tr trong 4 v tr cn li l ch s 7 nn c 4 cch chn. Ba v tr cn li l b phn phn bit th t c chn t E \ {1,7}nn c35Acch chn. Vy: s cc s gm 5 ch s phn bit hnh thnh t tp E trong c ch s 7 v ch s hng ngn lun l ch s 1 bng 1.4.35A= 240 s < 19 > t E = {1,2,3,4,5}a. * Gi s t nhin gm 5 ch s i 1 khc nhau l: 1 2 3 4 5a a a a a (10 a= ) a1 c 5 cch chn. a2 c 4 cch chn. a3 c 3 cch chn. a4 c 2 cch chn. a5 c 1 cch chn. Vy c : 5.4.3.2.1 = 120 s. * Gi s t nhin gm 5 ch s i 1 khc nhau, bt u bng ch s 1 l:1abcdth :a c 4 cch chn( v a E \ {1}) b c 3 cch chn( v b E \ {1, a}) c c 2 cch chn( v c E \ {1, a, b}) d c 1 cch chn( v d E \ {1, a, b, c}) suy ra c: 4.3.2.1 s bt u t ch s 1 . Vy:sccstnhingm5chskhc nhau,khng btubngchs 1l: 120 - 24 = 96 s. b.Gist nhingm 5chs bt u bngchs 123l:123xy gm 5chs khc nhau. x c 2 cch chn( v x E \ {1,2,3}) y c 1 cch chn( v y E \ {1,2,3,x}) = c 2.1 = 2 s. Vy: cc s t nhin gm 5 ch s khc nhau, khng bt u bng ch s123 gm c 120 - 2 = 118 s.