MT S DNG TON CC TR T HP, RI RC V NH HNG CCH GII

CC KIN THC C BN V T HP

QUI TC MQUI TC CNG( The Addition Priciples- AP):

Nu c

i tng khc nhau trong tp hp th nht , i tng khc nhau trong tp hp th hai , i tng khc nhau trong tp hp th m, Th th s cch chn 1 i tng t 1 trong m tp hp l .

Cch pht biu khc:

Cho l m tp hp hu hn , k1.Nu cc tp hp ny i mt ri nhau , ngha l th:.

QUI TC NHN (The Multiplication MP)Gi s c 1 qu trnh c th chia thnh m giai on lin tip nhau c th t , Vi k qu khc nhau trong giai on th nht , kt qu khc nhau trong giai on th hai , kt qu khc nhau trong giai on th m, nu cc kt qu kt hp li l phn bit Th th s kt qu kt hp li ca ton b qu trnh l .

Cch pht biu khc:

Cho l tch Decarste ca cc tp hp hu hn . Khi , ta c.

MT S V D:

1. C bao nhiu cch chn ra 4 s nguyn dng t tp hp S={1;2;.;499;500} sao cho l 1 cp s nhn tng v cng bi ca chng l mt s nguyn dng .

GII:Gi l 4 s cn chn , th th ta c:

Cho nn: .iu c ngha l s cp s nhn vi cng bi q l . Theo qui tc cng , s cp s nhn tha iu kin l:

.

2. C bao nhiu s t nhin l c 4 ch s phn bit?

GII:Mt s c 4 ch s l mt b sp th t ca 4 ch s ( ch s 0 khng ng u). V cc s cn m l cc s l nn ch s n v c th l 1,3,5,7,9. Ch s hng chc v trm c l 0,1,2,.9 v ch s hng ngn l 1,2,.9.V cc ch s l phn bit nn:-C 5 cch chn ch s hng n v . -C 8 cch chn ch s hng ngn.( khc 0 v khc ch s n v ).- C 8 cch chn ch s hng chc ( khc ch s n v v hng ngn).-C 7 cch chn ch s hng trm .Vy c 5.8.8.7= 2240 s cn chn.

3. Tm s cp c th t (x;y) ca cc s nguyn x,y sao cho .

GII:Ta phn chia bi ton thnh 6 trng hp ring bit:

Vi mi i=0;1;2;3;4;5 ta t .D kim tra:

Vy .

4. Tm s c s dng ca 600 , bao gm c 1 v 600

GII:Trc ht ta ch rng 600= . Khi 1 s nguyn dng m l c s ca 600 nu v ch nu m c dng: vi a,b,cZ:0a3; 0b1; 0c2. Nh vy s c s l 4.2.3= 24.M RNG: Nu mt s t nhin n c dng phn tch tha s nguyn t( The prime decomposition) l : trong l cc s nguyn t phn bit v . Th th s cc c s dng ca n l .

5. (AIME 1988)Tnh xc sut chn c ngu nhin 1 c s nguyn dng ca l mt bi s ca .

GII:c s ca c dng .C 100 cch chn a , 100 cch chn b, nn c 100100 c s ca .Tng t , bi s ca phi tha mn bt ng thc 88a;b 99, a,b Z; Nn c 12 cch chn a, v 12 cch chn b. Do xc sut s l: .

6. Xc nh s cc cp s c th t (a;b) sao cho bi chung nh nht ca a v b l .

GII:C a,b u l c ca nn

V l BCNN ca a,b nn max{x;s}=3; max{y;t}=7; max {z;u}=13. Bng cch lit k ta c 7 cch chn cp (x;s); 15 cch chn (y;t); 27 cch chn (z;u). Theo qui tc nhn , ta c 71527=2835 cp s (a;b) tha iu kin . M RNG:Nu n l s nguyn dng v l phn tch thnh tha s nguyn t ca n.Th s c cp s nguyn (a;b) phn bit c th t sao cho BCNN(a;b) l n.

7. Cho X={1;2..;100} v t S= . Tnh .

GII:Bi ton c th chia thnh cc trng hp phn bit khi xt a=1;2;.;99.Vi a=k{1;2;.;99} th s cch chn b l 100-k v c cng l 100-k , Th th s b s (a;b;c) cn tm l . V k ly gi tr 1,2,.,99 nn ta c :

8. Hy xc nh s hnh vung m cc nh ca n l cc im trong li vung 1010 sau (10 im , 9 ).

GII:Ta ni rng 4 im nn quartet ( nhm 4 ) nu chng l cc nh ca hnh vung nn m cc cnh ca n song song vi ng bin ca li. Ta cng ni rng 1 hnh vung vi cc nh ca 1 quartet l mt quartet square. Ta c 81= quartet 11. Ta c : 8 quartet 22 trong li 310 v c 8 li 310 trong li 1010 .Vy , c quartet 22 trong li 1010.Tng t ta c : quartet 33 trong li .Ngha l , khi k{1;2;;9} c quartet kk . Nhng phn kh khn l cc hnh vung c cnh khng song song vi ng bin ca li .Mi hnh vung ny s ni tip bn trong 1 quartet. Cho nn ta ch cn m tt c cc quartet v cc hnh vung ni tip n. Khng kh khn g ta c trong 1 kk quartet c k hnh vung ni tip , k c n . V d khi k=4 ta c hnh v bn.

Nh vy ta c:

9. C n que c di l 1,2,.n. C bao nhiu tam gic khng cn c to thnh t 3 trong s cc que ?

GII:Gi x,y,z l di 3 que . Khng mt tnh tng qut , ta gi s rng x z . Ta xt 2xz v 2x> z ngha l 1xm; v m z-x= k-x . V k=2m2x nn ta c k-x x( tha iu kin y>x). Cho nn bt k y nm gia k-x+1 v z-1= k-1 nh vy ta c (k-1)-(k-x+1) +1 =x-1 gi tr m y c th nhn c . Khi m2x >2m=z ( tha iu kin), Bt k y nm gia x+1 v k-1 nh th s c(k-1)-(x+1)+1= k-x-1= =2m-x-1 gi tr m y c th nhn c. Bi vy cho nn; khi k=2m

ngha l:

Ch rng cng thc ny vn ng khi m=1 ngha l khi k=2.Trng hp 2: Trong trng hp ny ta gi s k l , ngha l k=2m+1 vi k l s nguyn , m2. Khi 1xm , ta cng cn c y> z-x = k-x. Lc ny , k=2m+1>2x nh th k-x >x . Nh trc , y c th ly cc gi tr nguyn nm gia k-x+1 v k-1 , nh th s c (k-1)- (k-x+1) + 1= x-1 gi tr m y c th nhn . Khi m < x . Nh th s c (k-1)- (x-1)+ 1= k-x-1= 2m-x gi tr m y c th nhn c . Bi vy , cho nn khi k= 2m+1

.Ch rng cng thc vn ng khi m=0 v m=1 tc l k=1 v k=3.By gi ta bt u gii bi ton .Nu n l: n=2p+1. ( vi p l s nguyn khng m no )> Ta c:

Nu n chn: n=2p vi p l s nguyn dng . Th ta c:

Mt cch gii khc khi ta gii c n=2p+1 th trng hp n=2p ta c th tch nh sau:

10. Trc khi Rick m mt t ng th thao ca mnh , th Rick phi nh mt m kha ca t , Hai trong cc b 3 cp s ca mt m l 17 v 24 , nhng anh ta khng nh c cp s th ba. V khng nh c th t ca 3 cp s . C 40 kh nng ca cp s th ba. Trong 10 giy , th c th nh c tt c cc kh nng xy khng?

GII:Ta xt 6 tp con ca cc kh nng ca m kha .

Mi tp con c 40 phn t nhng phi tr i cc phn t chung ca 2 tp hp nh (17;17;24); (17;24;17); (24;17;17); (17;24;24); (24;24;17);(24;17;24). Vy c tt c 406-6= 234.

11. Mt bng li xe cha 1dy 3 k t alphbet theo sau l mt dy 3 ch s. C bao nhiu bng li xe c to thnh nu o v 0 khng dng cng 1 lc .

GII:Gi l tp hp cc bng li khng c s 0. l tp hp cc bng li khng c ch o. l tp hp cc bng li khng c s 0v khng c ch o.Ta c:

12. Xc nh s s nguyn dng nh hn 1000 cha t nht mt ch s 1 trong cch vit thp phn.

GII:CCH 1:Gi S l tp hp cc s nguyn dng nh hn 1000.

Gi l tp hp cc s nguyn dng c 1 , 2, 3 ch s.Vi i=1;2;3 t cha ng cc ch s v c t nht 1 ch s 1.Ta ch cn tinh .Ta d dng thy rng .Ta c th chia thnh 3 tp hp: .Ta c th chia

. Vi a,b,c1;a0.Do : .Nh vy ta c =271.

CCH 2:Ta c th phn chia tp hp nh sau: Gi S l tp hp nhng s nguyn khng m nh hn 1000. Gi l tp hp cc s nguyn khng m nh hn 1000 cha t nht 1 ch s 1, v Gi l tp hp cc s nguyn khng m nh hn 1000 khng cha ch s 1. Ngha l

Ta c Nn .

13. C 15 l thng hi my lnh trong 1 rp ht . gi cho nhit mt m , phi c t nht 1 l thng hi lm vic sut thi gian. Hi c bao nhiu cch thc hin.

GII:Ta gi cc l thng hi l Ta c mi l thng hi c 2b la chn l on hay off . Ta s dng k t n cho on, v f cho off. Ta nh du 1 m gm 15 k t cho mt tnh hung. Nh vy theo qui tc nhn c tnh hung , nhng tnh hung fffff.f b loi v tt c cc l thng hi u ng . Do c =32767 tnh hung xy ra.

Gi A v B l 2 tp hp . Mt nh x f ( map- mapping function) t tp hp A n tp B ( vit l f:A B). nh du mi phn t aA vi ng mt phn t bB ( vit l f(a)=b). b l nh ca a. Vi AA , Gi f(A) ( nh ca A) xc nh tp hp cc nh ca phn t aA. Nu f(A)= B th f c gi l ton nh (surjective- onto)ngha l , vi mi bB s l nh ca aA. Vi mi 2 phn t phn bit c nh khc nhau th f c gi l n nh ( injective- one to one) . Nu f va l n nh va l ton nh th f l song nh( bijective one-to-one correspondence) HON V : ( PERMUTATION)(khi nim m rng):tng ng vi khi nim chnh hp ca sgk)Mt cch sp xp th t m phn t phn bit ca n phn t phn bit cho trc ( mn) c gi l 1 hon v ly m phn t ca n phn t . V cc phn t l khng lp li nn nn hon v cng l khng lp v s hon v ly m phn t ca n phn t phn bit c k hiu l .Khi ta c :.( mn).c bit l khi m=n th c gi l hon v ca n phn t, S hon v ca n phn t l

14. Cho tp hp E={a;b;c;;x,y,z} l tp hp 26 ch ci ting Anh. Tm s t c 5 k t sao cho cc k t c to thnh t tp E; v k t u v cui l cc nguyn m phn bit , cc k t cn li l cc ph m phn bit.

GII:p s : .

15. Trong 1 cuc hp c 7 nam v 3 n . C bao nhiu cch sp xp h thnh 1 hng sao cho :a/ 3 ngi n to thnh 1 nhm ( ngha l khng c ngi nam no ngi gia 2 ngi n )b/ Hai v tr hai u l nam v khng c n no ngi k nhau.

GII:A/ p s : 8!.3!.B/ Trc ht ta sp xp 7 nam : c 7! cch xp.Vi mi cch xp c nh , v v tr 2 u l nam , nn ch c 6 ch cho 3 n . Vy s cch xp l 7!654.

16. Gia cc s 20000 v 70000 , tm s cc s nguyn chn m cc ch s khng lp li.

GII:Gi abcde l s cn tm , Ch s a c th chn t {2;3;4;5;6} v ch s e c th chn t {0;2;4;6;8} . V {2;3;4;5;6} {0;2;4;6;8}={2;4;6} nn ta chia thnh 2 trng hp ri nhau:Trng hp 1: a{2;4;6} : a c 3 cch chn , e c 5-1=4 cch chn , v bcd c cch chn .Vy c s chn.Trng hp 2: a{3;5}a c 2 cch chn , e c 5 cch chn v cn li bcd c cch chn . Vy c s chn.Vy c tt c 7392 s chn.

17. Cho S l tp hp cc s t nhin m cc ch s c chn t {1;3;5;7}. Sao cho khng c ch s no lp li. Tm a/

b/

GII:CCH 1:a/ Ta chia S thnh 4 tp hp con ri nhau:(1) cc s c 1 ch s: 1;3;5;7.(2) cc s c 2 ch s : 13;15;17;(3) cc s c 2 ch s : 135;137;(4) cc s c 4 ch s: 1357;1375;..Ta c : .b/ Gi xc nh tng ca cc ch s hng n v ca cc s trong S,

xc nh tng ca cc ch s hng chc ca cc s trong S;

xc nh tng ca cc ch s hng trm ca cc s trong S

xc nh tng ca cc ch s hng ngn ca. cc s trong S.Do ;

Trc ht , ta xc nh ; R rng , tng ca cc ch s hng n v trong l 1+3+5+7 =16. Trong c s m cc ch s n v tng ng l 1,3,5,7 nn tng cc ch s hng n v ca cc s trong , l .Trong c s m cc ch s n v tng ng l 1,3,5,7 nn tng cc ch s hng n v ca cc s trong , l .Trong c s m cc ch s n v tng ng l 1,3,5,7 nn tng cc ch s hng n v ca cc s trong , l .Cho nn :

Tng t , ta c :

Vy

.

CCH 2:R rng 4 s trong c th chia thnh 2 cp {1;7} v {3;5} v tng ca 2 s trong mi cp ny l 8. 12 s trong c th chia thnh cp {13;75};{15;73};{17;71};{35;53} v tng ca 2 s trong mi cp ny l 88. Tng t ; 24 s trong c th chia thnh cp v tng ca 2 s trong mi cp ny l 888. 24 s trong c th chia thnh cp v tng ca 2 s trong mi cp ny l 8888. Nh vy :

T HP ( COMBINATION)

18. Chng minh cng thc : (n,rN; rn).

GII:Cch 1: Dng cng thc.Cch 2: Dng T hp . Gi A={1;2;;n} c cch ly r phn t t A. Mi cch ly s c phn t 1 hoc khng c phn t 1. S cch ly c phn t 1 l . ( ch cn ly thm r-1 phn t) S cch ly khng c phn t 1 l . T suy ra pcm.

Mt dy cc s c gi l xu k-aray , vi n,k , nu vi mi i=1;2;;n. di ca xu l n , chnh l s s hng trong xu .i khi mt xu nh th c th vit l . Mt xu k-aray c th c gi l xu nh phn ( binary) , Xu tam phn ( ternary) hay t phn ( quarternary) khi k=2 , 3, 4. V d: {(0;0;0); (0;0;1);(0;1;0);(1;0;0);(0;1;1);(1;0;1);(1;1;0);(1;1;1) }l tp hp tt c 8 xu nh phn c di l 3. to ra 1 xu k-aray c di n: trc ht ta chn t tp hp B={0;1;;k-1} k n chn t tp hp B={0;1;;k-1}; v cho n cui cng chn t tp hp B={0;1;;k-1},V c k cch chn trong mi bc , nn s xu k-ary phn bit c di n l

19. Cho dy nh phn c chiu di l 7.Hi c bao nhiu dy cha 3 s 0 v 4 s 1.

GII:Trc ht ta xp 3 s 0 vo 3 trong 7 v tr ca chui .Sau xp 4 s 1 vo 4 v tr cn li.Vy c chui tha iu kin .

20. C bao nhiu cch lp ra 1 y Ban gm 5 ngi t 11 ngi bao gm 4 thy gio v 7 hc sinh, nu :a/ Khng c yu cu v cch la chn .b/y ban phi bao gm ng 2 thy gio .c/ y ban phi bao gm t nht 3 thy gio.d/ c bit 1 thy gio v 1 hc sinh khng th cng nm trong y ban.

GII:a/

b/ .c/.d/ Gi T l ngi thy c bit, S l hc sinh c bit. Ta tm s cch lp ra y ban bao gm c T v S. Nh vy c cch lp . Vy s cch lp tha iu kin l

21. Gi s c 8 ngi chi a;b;c;d;e;f;g;h tham d 1 gii tennis n . vng u tin, h chia thnh 4 cp thi u. Hi s cch sp xp.

Gi A l 1 tp hp gm 2n phn t . Mt ghp i ca A ( A pairing ) l 1 s phn chia tp hp A thnh cc tp con 2 phn t ri nhau tc l 1 hp ca cc tp con 2 phn t ri nhau to thnh A. Th d : Nu A={a;b;c;d;e;f;g;h} th {{a;b};{c;f};{d;h};{e;g}} v {{a;h};{c;f};{d;h};{b;g}} l 2 ghp i khc nhau ca A.Ch rng th t ca cc tp con v th t ca 2 phn t trong mi tp con l khng quan trng .

22. Cho A l tp hp gm 2n phn t ( n1) Tm s s ghp i khc nhau ca A.

GII:Ta xt 3 cch khc nhau gii bi tp ny :Cch 1: Ta chn mt phn t bt k l x trong A. S cch chn ngi cng cp vi x , gi l y, l 2n-1 cch . ( v {x;y} l 1 tp con 2 phn t).Chn mt phn t bt k khc l z, t 2n-2 phn t cn li ca tp hp A\{x;y} . S cch chn ngi cng cp vi z l 2n-3. Tip tc qu trnh .S cch cn tm l :(2n-1)(2n-3)..5.3.1.Cch 2:Trc ht ta to ra 1 tp con c 2 phn t ca A v t vo v tr (1) nh hnh v .C cch l nh th .

K tip , li to ra 1 tp con c 2 phn t t phn cn li ca A v t chng vo v tr (2) . C cch lm nh th . V c tip tc nh vy.Do s cch cn tm l :v cc phn chia khng quan tm n th t.

Cch 3: Trc ht ta sp xp 2n phn t ca A thnh 1 hng v t chng vo 2n khong trng sau: { ; };{ ; };.{ ; }; (1) (2) (3) (4) (2n-1) (2n)C (2n)! cch sp xp . V th t ca cc phn t trong mi tp con 2 phn t v th t cc phn t trong n tp con l khng quan trng nn s cch theo yu cu l : .

CH :Bi ton trn c th m rng theo cch sau: Cho A l mt tp hp c kn phn t phn bit (k,nN) . 1 s ghp k-phn t ca A l mt phn hoch A thnh cc tp con k phn t tc l phn chia A thnh cc tp con k phn t i mt ri nhau.

23. C bao nhiu s c 5 ch s ln hn 21300 sao cho cc ch s ca n l phn bit v ly t cc ch s {1;2;3;4;5}

GII:Cch 1:Ta chia thnh cc loi:- S cc c 5 ch s m ch s hng chc nghn c th l 1 trong cc s 3,4,5: l.- S cc c 5 ch s m ch s hng chc nghn c th l s 2 v ch s hng nghn l 1 trong cc s 3,4,5 l .-S cc c 5 ch s m ch s hng chc nghn c th l 2 v ch s hng nghn l 1 l .Vy tng s cc s l .Cch 2:V s cc s c 5 ch s m cc ch s phn bit l v ch c cc s m ch s hng chc ngn l 1 th mi khng vt qu 21300 ( s cc s ny l ) nn s cc s cn tm l .

24. Cho n,k l cc s nguyn dng v S l tp hp n im trong mp sao cho:(i) khng c 3 im no ca S l thng hng ,(ii) Vi bt k im P thuc S , c t nht k im thuc S cch u P.Chng minh rng

IMO 1989.

GII: thun li , ta gi 1 on thng ni 2 im bt k ca S l cnh . Gi l l s cnh trong mp. Trc ht , v c n im phn bit v bt k 2 im xc nh c 1 cnh nn ta c l=cnh . K tip , mi im P ca S theo iu kin (ii) c th v c 1 ng trn tm P(C(P)) sao cho ng trn cha t nht k im ca S. R rng rng mi im ca S nm trn (C(P)) xc nh t nht cnh . Do vi n im P trong tp hp S th tng s cnh t nht l n ( c m lp li). By gi ta thy rng , cc cnh c m nhiu hn 1 ln. 1 cnh c m nhiu hn 1 ln khi v ch khi cnh l dy cung chung ca t nht 2 ng trn. V 2 ng trn c nhiu nht 1 dy cung chung cho nn n ng trn , s dy cung chung c m lp li nhiu nht . Cho nn :

.

NHN XT:(1) Trong chng minh trn , i lng l c dn ra t u v n c m c lng bng 2 cch suy ngh, bng cc cch y dn n bt ng thc cn chng minh. l 1 k thut hay s dng trong t hp . (2) T chng minh trn ta thy rng iu kin (i) l khng cn thit v nu A,B,C thng hng th 3 on thng AB,AC,BC cng c xem nh l 3 cnh phn bit.

HON V LP LI :

Mt cch sp xp m phn t ca n phn t phn bit ( mi phn t c th lp li hu hn ln )c gi l hon v lp li ca m phn t t n phn t.S hon v lp li l .

Chng minh :Xc nh n phn t phn bit l 1,2,3n. Th th mt t hp lp c dng :

t

v nh vy mi t hp lp s tng ng vi duy nht mt t hp khng lp v s t hp ny l .

ALL PERMUTATION OF INCOMPLETE DISTINCT OBJECTS:Gi s c n phn t bao gm k phn t phn bit vi s ln lp li tng ng l , tt c cc hon v ca n phn t c gi l tt c cc hon v ca cc i tng phn bit khng y , ta c s hon v loi l .Chng minh :Gi f l s hon v tha mn bi ton .Nu ta trao i cc phn t trong cng mt loi v sp xp ln nhau tng nhm th ta s c hon v . Theo qui tc nhn s tt c cc hon v ca n phn t phn bit bng .Do ta c : f.= n! Suy ra pcm.

25. Tm s dy tam phn c di l 10 c 2 ch s 0, 3 ch s 1v 5ch s 2.

GII:p s : 2520.

26. Tm s cch lt 1 hnh ch nht 17 bng cc block 11; 12; 13. Gi s rng cc block ny c cng kch thc l khng ng k.

GII:Minh ha 2 cch ph nh sau:

2131

133

Vi i =1 ;2;3 , ta t xc nh block 1i. Nh vy , cch lt th nht ch ra trn c th biu din di dng . l hon v ca b . Cch th hai biu din bi . l hon v ca .Ch rng tng ca cc ch s trong mi b u bng 7.GII:T minh ha trn , ta thy s cch yu cu bng vi s hon v ca mt vi s sao cho tng ca cc ch s ca l 7. Ta c 8 trng hp bao gm cc kh nng sau y:

Trong mi trng hp , s hon v l :

Nh vy s c : 44 cch lt .

27. Chng minh rng (4n)! l bi s ca vi mi s t nhin n.

GII:Xt tp hp .Ta c : .V l s t nhin nn suy ra pcm.

T HP LP LI:

Mt la chn khng th t m phn t t n phn t cho phn bit ( mi phn t c th lp li hu hn ln ) c gi l t hp lp . S t hp lp l .Cho M= l 1 multi-set vi n N.Mt multi-set ca dng vi l cc s nguyn khng m, c gi l mt -phn t ca multi-subset cua M. Vi s nguyn khng m r , goi xc nh s r-phn t ca multi-subset ca M. 28. C 3 loi sandwich , gi l chicken(C ) ; fish(F) ; ham (H). c nh hng. Mt ngi mun t trc 6 sandwich . Gi s rng khng gii hn vic cung cp sandwich trong mi loi , Hi c bao nhiu cch t phn n trn?

GII:Vn ca bi ton ny l m s c lit k trong 6-phn t multi subset ca . Bng sau y cho 4 cch la chn :

(C)(F)(H)

(1)o o o o o o

(2)o o o o o o

(3)o o o o o o

(4)o o o o o o

Ta biu din bng trn bng chui nh phn :(1) 00101000(2)01000010(3)10010000(4)00011000Trong mi trng hp , ta thy vic sp th t ca 6 loi sandwich tng ng vi mt chui nh phn c di l 8 vi 6 k t 0 v 2 k t 1. , th t khc nhau dn n chui nh phn khc nhau. Ta thy c 1 song nh gia tp hp cc cch sp xp vi tp hp chui nh phn nh trn. Cho nn ta c , .Xt trng hp tng qut:

S

S dng c cu ny , mi tp con S=ca M , y c th biu din thnh 1 hng c s 0 trong khong di . Nu ta xem mi ct ng l 1 s 1 th mi r-phn t tp con ca M tng ng vi t chui nh phn c di r+n-1 vi r s 0 v n-1 s 1 . Tng ng ny l mt song nh gia h cc r-phn t tp con ca M v h cc chui nh phn nh trn. Nh vy ta c kt qu,Cho M= l 1 tp con vi n N.S r-phn t ca M l . NHN T HP: Sp xp n phn t phn bit vo k loi phn bit ( kn) sao cho c phn t trong nhm th i , ( i=1;2.;k; ) Th th s cch sp xp l :

.Chng minh :S cch ly t n phn t l

S cch ly t n- phn t l

.

S cch ly t n phn t l

S dng qui tc nhn ta c s cch thnh lp l :

29. Gi s c 3 c , 4 c xanh v 2 c vng t vo v tr ca 9 ct c nh s . Hi c bao nhiu k hiu phn bit t cc cy c .

GII:S k hiu phn bit l : .

30. C bao nhiu cch chn ra 3 cp i t n ngi ( n6).

GII:Cch 1:S cch ly 6 ngi t n ngi : .6 ngi ny phn chia thnh 3 nhm , mi nhm c ng 2 ngi , s cch chia l : nhng 3 nhm ny khng cn th t nn s cch chn theo yu cu l:

Cch 2:

S cch ly 6 ngi t n ngi : .Vi 6 ngi , c chn ra 2 ngi, vi 4 ngi cn li c cch chn ra 2 ngi . Cui cng c chn ra 2 ngi cn li , Nhng 3 cp ny khng c th t nn s cch chn l :

HON V VNG TRN CA CC PHN T PHN BIT:Nu ta sp xp n phn t phn bit trn 1 ng trn th hon v c gi l hon v vng trn ca n phn t, S hon v vng trn ca n phn t l : .

CHNG MINH :V n hon v ng thng cho ta 1 hon v ng trn m ta li c n! hon v ng thng nn suy ra pcm.Cc hon v nhn c t nhau qua 1 php quay quanh tm , c xem l mt. Gi A l mt tp hp gm n phn t phn bit . Vi 0rn , s hon v vng trn r phn t t tp A c k hiu l . Ta chng minh c .

31. C bao nhiu cch 5 nam v 3 n ngi chung quanh 1 ci bn , nu :a/ nu khng c hn ch g ?b/ Nam B1 v n G1 khng ngi k nhau.c/ khng c n no ngi k nhau.

GII:a/ p s : 7!b/ 5 nam v 2 n ( khng tnh G1) c th c (7-1)! Cch xp .G1 c 5 cch ngi khng k vi B1. Vy c 6!5=3600 cch xp .Ta cng c th dng nguyn l Phn B:(PRINCIPLE OF COMLEMENTATION-CP)Nu A l tp con ca X th

S cch xp 5 nam v 3 n trong B1 v G1 ngi cnh nhau l : 6!.2!=1440.Vy s cch xp tha mn bi ton l : 7!- 1440=3600.c/ Trc ht ta xp 5 nam vo bn , c (5-1)! Cch xp.Ln lt c 5 cch xp G1; 4 cch xp G2; v 3 cch xp G3.Vy c tt c 4!543=1440 cch xp.

32. Tm s cch xp ch ngi cho n cp v chng xung quanh 1 bn trn sao cho :a/ Nam v n ngi lun phin.b/ Mi N ngi k vi chng ca mnh .

GII:a/ Xp n ngi nam c (n-1)! cch xp. n ngi n c th ngi vo n khong trong gia 2 ngi nam , nn c n! cch xp . Vy c n!.(n-1)! Cch xp.b/ Mi cp v chng xem nh 1 phn t .S cch xp n phn t ny l (n-1)! . V mi cp v chng c 2 cch xp , nn s cch xp tng cng l : .

CH : Mt bi ton kh hn v ni ting lin h vi bi tp trn l:C bao nhiu cch xp ch ngi cho n cp v chng (n3) quanh 1 bn trn sao chon nam v n ngi xen k nhau nhng v khng ngi cnh chng ?Bi ton ny ln u tin c gii thiu bi nh ton hc Php Francis Edward Anatole Lucas (1842-1891)

33. Nu phi xp t nht 1 ngi trn mt bn th c bao nhiu cch xp 6 ngi ngi:a/ quanh 2 bn .b/quanh 3 bn.

GII:a/ Vi 2 bn, c 3 trng hp xt s ngi ngi quanh 2 bn tng ng , (1) 5+1 (2) 4+2

(3) 3+3.Trng hp 1: C cch chia 6 ngi thnh 2 nhm vi s lng 5 v 1 mi bn. S cch xp 5 ngi vo 1 bn l (5-1)! v 1 ngi vo bn cn li l 0!. Theo qui tc nhn , ta c : .Trng hp 2: C cch chia 6 ngi thnh 2 nhm vi s lng 4 v 2 vo mi bn. S cch xp 4 ngi vo 1 bn l (4-1)! v 2 ngi vo bn cn li l 1!. Theo qui tc nhn , ta c : .Trng hp 3: Ta ch trng hp ny . S cch chia 6 thnh 2 nhm 3 v 3 l . Vy s cch xp l .Cho nn theo qui tc cng l : 144+90+40=274.

b/ Vi 3 bn ta c 3 trng hp : (1)4+1+1 (2)3+2+1 (3) 2+2+2.S cch sp xp trong mi trng hp l:

Theo qui tc cng : 90+120+15=225.

CH :Cho r,n l cc s nguyn , 0nr , xc nh s cch sp th t r phn t phn bit quanh n ng trn ( khng phn bit) sao cho mi ng trn c t nht mt phn t. ( k hiu l s(r;n)).Cc s s(r;n) c gi l s Stirling loi 1, mang tn nh ton hc James Stirling ( 1692- 1770) V d : s(6;2)= 274 ; s(6;3)= 225. Cc kt qu khc :s(r;0)= 0 nu r1 .s(r;r)=1 nu r0.s(r;1)= (r-1)! nu r2 .s(r;r-1)= nu r2.

34. Chng minh rng : s(r;n) = s(r-1;n-1) + (r-1)s(r-1;n) vi r,nN;nr.

GII: n gin , ta gi r phn t phn bit l 1,2,.r. Xt phn t 1 , vi bt k cch sp xp cc phn t , hoc l :(i) 1 ch l 1 phn t trong 1 ng trn .(ii) 1 trn vi cc phn t khc trong 1 ng trn.Trong trng hp th nht c s(r-1;n-1) cch xp .Trong trng hp th hai c s(r-1;n) cch xp cc phn t 2;3;r vo n ng trn , khi 1 c th t vo 1 trong r-1 khong trng phn bit n immediate right ca r-1 phn t tng ng. Theo qui tc nhn , trng hp ny c (r-1).s(r-1;n) cch sp xp. Suy ra pcm.

S dng cc gi tr ban u: s(0;0)=1 ; s(r;0)=0 ; s(r;1)= (r-1)! Vi r1ta c th tm c cc gi tr s(r;n) vi r,n kh nh.

S XU CHUI HT:Gi s 1 xu chui ht bao gm n ht c sp xp trn 1 ng trn th th s xu phn bit l 1 ( nu n=1 ;2) hay .CHNG MINH : Nu n=1 hay n=2 th s xu chui l 1. Gi s n 3, bi v 1 xu chui c th quay hay lt ngc li m khng lm thay i g , nn s xu chui bng s hon v vng trn. 35. C bao nhiu cch sp xp 6 n v 15 nam nhy ma theo vng trn sao cho c t nht 2 ngi nam ng gia bt k 2 ngi n .

GII: Trc ht vi mi ngi n , ta coi nh 2 ngi bn nhy nam ca c y l 1 ngi ng bn tri v 1 ngi ng bn phi. V c 6 ngi n phn bit nn ta c th chn 12 ngi nam t 15 ngi nam bng cch . K n , mi ngi n v 2 bn nhy nam ca mnh c xem nh l 1 nhm , mi phn d li 15-12=3 ngi nam cng xem nh l 1 nhm . Nh vy tng cng c 9 nhm , m ta c th hon v vng trn nn s c 8! Cch . Theo qui tc nhn ta c : cch .

S NGHIM CA PHNG TRNH V NH BC NHT.(bi ton chia ko ca Euler)S nghim nguyn khng m ca phng trnh bng

.CHNG MINH :Ta xt mi nghim khng m ca phng trnh l tng ng vi 1 hon v ca n ng trn O v m-1 cnh / :

. y l s ng trn O bn tri du / th nht, l s ng trn O gia du/ th i v du / th i+1, . l s ng trn O bn phi du / th m-1 . V tng ng trn l 1-1 , nn s nghim ca phng trnh bng s hon v ca n ng trn O v m-1 du/ tc l .

CH : S nghim nguyn khng m ca phng trnh bng s t hp lp li ca vic ly m phn t t n phn t ( mi phn t c th c hu hn ln lp li).H QU : S nghim nguyn dng ca phng trnh (1)bng .CHNG MINH :t Khi ta c : .(2)S nghim dng ca (1) bng s nghim khng m ca phng trnh (2) do s bng .

Li gii khc ca VD7: Gi s 15 nam chia thnh 6 nhm sao cho mi lnh o ca nhm l mt ngi n v c t nht 2 nam trong 1 nhm . Xc nh s nam trong mi nhm l tng ng , th th ta c phng trnh :

(1)t (2)Th th s nghim nguyn ca (1) bng s nghim nguyn khng m ca (2) tc l .Nh vy 15 nam c chia thnh 6 nhm sao cho trong mi nhm c t nht 2 nam th c cch. Ta sp xp 6 nhm trong vng trn th c (6-1)!=5! cch. Ngi lnh o ca mi nhm l 1 n v v tr ca n c xc nh . 15 nam ngi trn ng trn c 15! cch . Theo qui tc nhn ta c s hon v tha mn yu cu bi ton l .

36. C bao nhiu s nguyn c 3 ch s sao cho tng cc ch s ca mi s nguyn l 11.

GII:Ta xc nh ch s hng trm , hng chc v hng n v l : (1).t (2) .S nghim ca (1) bng s nghim ca (2) tc l bng nhng tr i 5 nghim sau : (11;0;0) ; (10;1;0) ;(10;0;1); (1;10;0); (1;0;10) vy c tt c :

NGUYN L BAO HM V NGUYN L LOI TR :Gi l n tp hu hn . Ta xc nh s phn t ca l . Th th ta c cng thc (1) sau:

CHNG MINH :Vi bt k phn t x ta chng minh rng x s c s ln m ging nhau c 2 v ca (1).V x thuc t nht 1 trong cc tp hp nn khng mt tnh tng qut ta gi s x v khng thuc cc tp hp khc . Trong trng hp ny x c m 1 ln trong VT ca (1) . Nhng trong VP ca (1) x c m ln trong ; c m ln trong ; c m ln trong ; . Suy ra , VP ca x c m :

R rng rng vi x th c 2 v x u khng m c ln no. Nh vy 2 v mi phn t x c m s ln nh nhau , nn cng thc c chng minh.

CH : Phng php chng minh trn c gi l phng php GP LI ( CONTRIBUTED METHOD).

NGUYN L QUT LIN TIP ( SUCCESSIVE SWEEP PRINCIPLE):SIEVE FORMULA CNG THC SNG :Cho S l tp hp hu hn , (i=1;2..;n) v xc nh phn b ca trong S l (i=1;2;.;n). Th th :

CHNG MINH :Ta c :theo cng thc De Morgan

37. Xc nh s cc s nguyn dng nh hn 1000 v khng chia ht cho 7 v cng khng chia ht cho 5.

GII:t S={1;2;;999} Th th ta cn tnh : .Theo cng thc sng , ta c:

38. C bao nhiu cch gn n l th phn bit vo trong n phong b sao cho khng c l th no vo ng phong b tng ng ca n ( BI TON CC L TH SAI A CH CA BERNOULLI-EULER)

GII:Ta pht biu li bi ton : C bao nhiu hon v ca cc s {1;2;;n} sao cho s k khng t v tr k vi mi k (1kn). Cc hon v nh th gi l hon v xo trn ( derangement) v s hon v nh th t l .Gi S l tp hp cc hon v ca {1;2;;n} v l tp hp cc hon v ca {1;2;;n} tha iu kin (i=1;2;.;n).R rng ta c :

Theo cng thc sng , ta c:

HON V V CC IM BT NG CA N:Cho X={1;2;;n} ; l song nh t X vo chnh n v ta thng vit hon v di dng sau y:

Vi iX, nu , th i c gi l 1 im bt ng ca hon v trn X. Vi VD trn , ta c h qu sau y:H QU : S hon v khng c im bt ng ca tp X bng

39. Gi s tp X={1;2;;n} v xc nh s hon v khng c im bt ng ca X l , s hon v c ng 1 im bt ng ca X l . Chng minh rng .

GII:Gi l s hon v vi ng 1 im bt ng l I (i=1;2.;n) Th th :

Theo h qu trn , ta c :

. Nn :

40. Mt dy s c nhn t dy cc s nguyn dng {1;2;3; } bng cch xa i tt c cc bi s ca 3 hay 4 ngoi tr 5. Tinh .

GII:CCH 1: ( PHNG PHP C LNG GI TR- ESTIMATE VALUE METHOD) t ; S={1;2;3.;n} v (i=3;4;5) .Th th tp hp cc s khng b xa l .p dng cng thc sng , ta c :

p dng bt ng thc , ta c :

Nh vy ta c : .Nu n l bi s ca 3 hay 4 nhng khng phi 5, th n khng l s hng ca dy s mi , nh th theo yu cu n ch l 1 trong cc s sau: 3345;3346;3347;3349; 3350;3353.Ln lt th cc s trn vo phng trnh ban u ta c n= 3347.Vy .

CCH 2: PHNG PHP PHN TCH T HP( COMBINATORIAL ANALYSIS METHOD):Ta c BCNN(3;4;5)= 60 . t

Th th tp hp cc s khng b xa trong . p dng cng thc sng , ta c:

Do c 36 s hng trong dy s mi trong l :

Gi v

V hay nhng

Mt khc vi bt k s nguyn dng c dng 60k+r ( k;r l cc s nguyn khng m v rP) . Nu (r;12)=1 th (60k+r;12)=1 , nh th 60k+r cng l s hng ca dy s mi . Nu (r;12)1 th 5/r (v rP) , nh th 5/60k+r th 60k+r cng l s hng ca dy s mi . Nh th , dy s mi bao gm tt c cc s dng c dng 60k+r ( k,r l cc s nguyn khng m v rP) . Vi k cho trc , ta nhn c 36 s hng lien tip ca dy s mi . Ch rng 2009= 3655+29 nh th . Nhng

.Vy

BI TP P DNG :

1/ Mt thy gio trao n+1 gii thng cho n hc sinh sao cho mi hc sinh nhn t nht 1 gii thng . Hi c bao nhiu cch trao gii?

GII:Theo iu kin , phi c 1 hc sinh nhn 2 gii thng.v mi hc sinh cn li nhn mt gii thng. p s : D.2/Gi s mt thy gio chn 4 sinh vin t 5 nam v 4 n lp ra 1 nhm tho lun , Hi c bao nhiu cch thnh lp nhm tho lun c t nht 1 nam , 1 n . a/60

b/80

c/120 d/420.GII:S cch chn l :

p s : C.3/Nu 1 s c 5 ch s ln hn 20000 khng l bi s ca 5 c tnh cht sau: cc ch s ca n l phn bit v mi ch s l mt trong cc s 1,2,3,4,5 th th s s cn tm l a/96

b/76

c/72 d/36.GII:Trc ht , ch s n v khng th l 5. Vy s cc s c 5 ch s l 4!4. Trong cc s , s cc s c 5 ch s khng qu 20000 l ch s u l 1 v ch s n v l 1 trong cc s 2,3,4. Vy c 33!. Vy s s cn tm l : 44!-33!= 78.p s : B4/ Nu cc h s A v B ca phng trnh ng thng Ax+By =0 l 2 ch s phn bit t s 0;1;2;3;6;7 th s ng thng phn bit l bao nhiu?GII:S cch chn 2 s A v B trong 6 ch s l : .Nhng trong tnh ton trn , ng thng trong cc trng hp sau cng l mt ng thng : (1) A=0 v B=1;2;3;6;7.(2) B=0 v A=1;2;3;6;7.(3) A=1 v B=2 hoc A=3 v B=6.(4) A=1 v B=3 hoc A=2 v B=6.(5) A=2 v B=1 hoc A=6 v B=3.(6) A=3 v B=1 hoc A=6 v B=2.Cho nn s ng thng phn bit l :

p s : 18.5/Nu c s a v bin s x trong l 2 ch s phn bit t cc ch s 1;2;3;4;5;7;9 th s gi tr khc nhau ca

GII:S cch chn 2 s a v x t 7 s l : .Nhng c s a khng th bng 1.V cc trng hp sau trng lp nhau:(1) x=1 v a=2;3;4;5;7;9.(2) .(3) .(4).(5)

Nh vy s trng hp cn tm l :

6/ Trong 1 cuc thi u tennis , mi ngi chi ng 1 game vi mi ngi khc . Nhng trong qu trnh thi u , c 3 ngi b cuc v mi ngi tham gia ng 2 trn . Nu tng s trn l 50 th s trn ca 3 ngi trn l bao nhiu CHINA MATHEMATICAL COMPETITION 1994a/0

b/1

c/2 d/3

GII:Gi s c n ngi chi , v s trn 3 ngi chi l r . Th th :

Vy r=1.p s : B.

7/ Gi s a,b,c trong phng trnh ng thng ax+by+c=0 l 3 phn t phn bit trong tp hp {-3;-2;-1;0;1;2;3} v h s gc ca ng thng l dng . Hi c bao nhiu ng thng phn bit ? CHINA MATHEMATICAL COMPETITION 1999.GII:Gi s h s gc ca ng thng l .Khng mt tnh tng qut , gi s rng a>0 v b