83
(Giii
MUl til 1, Gpi so'can lap X = abed , a,b,c,d e |0,l,2,3,4,5,6};a * 0
Chon a: c6 6 each; chon b,c,d c6 6.5.4 Vay CO 720 so.
2. Gpi x = abcde l a s o c a n l a p , e € { 0 , 5 } , a ; t 0 ^^:rc,^.;.h-:';fi fy.y.y..
• e = 0 e CO ] each chon, each chon a,b,e,d : 6.5.4.3 ^ <• J^- T r u o n g h o p nay CO 360 so'. l^U 'A' '
• e = 5 => e CO m o t each chon, so each chon a,b,c,d : 5.5.4.3 = 300 T r u o n g h o p nay co 300 so.
Vay CO 660 so thoa man yeu cau bai toan. •
V i d y 8. Cho tap hop .so: A = {0,1,2,3,4,5,6}. H o i c6 the thanh lap bao nhieu so CO 4 ehii so'khac nhau va chia he't cho 3.
G i a i '^^-^ ^ ^ "
Ta CO m o t so chia he't cho 3 k h i va chi khi tong cac chu so chia he't cho 3.
T r o n g tap A c6 cac tap eon cac chi> so chia he't cho 3 la (0,1,2,3}, (0,1,2,61, (0,2,3,41, (0,3,4,51, (1,2,4,51, (1,2,3,61, { l , 3 , 5 , 6 } . , , , , Vay so cac so can lap la: 4 ( 4 ! - 3 ! ) +3.4! = 144 so. - A ^^int .t V i d u 9. T u cac so eua tap A = {0,1,2,3,4,5,6} c6 the lap dupe bao nhieu so c h i n gom 5 chi> so d o i m o t khac nhau trong do c6 hai c h u so le va hai c h u s o l e d u n g canh nhau.
V i CO 3 so le la 1,3,5, nen ta tao dupe 6 cap so kep: 13,31,15,51,35,53 Gpi A la tap cac so g o m 4 chu so dupe lap t u X = {0,13,2,4,6 .
Gpi A i, A 2 , A 3 t u o n g u n g la so cac so t u nhien le gom 4 c h u so khac nhau dupe lap t u cac c h u so cua tap X = {0,13,2,4,6} va 13 d u n g 6 vj t r i t h i i nhat, t h u hai va t h u ba. J^M•a•>?y:J ^'^J^ r, r ! f,i' > : v A - , , '
Taco: |A,| = A^ = 24;|A2| = [Ag] = 3.3.2 = 18 nen |A| = 24 + 2.18 = 60 Vay so cac so can lap la: 6.60 = 360 so.
V i d u 10. T u cac so 1,2,3,4,5,6 co the lap dupe bao nhieu so t u nhien ,m6i so c6 6 chii so dong thai thoa man dieu ki^n: sau so ciia moi so la khac nhau va trong m o i so do tong ciia 3 ehi> so dau nho han tong ciia 3 so sau mot d o n vj.
84
Cty TNHH MTV DWH Kluutg Vict
G i a i i
* Cach 1: Gpi X = 3^32...ag, a| €{1,2,3,4,5,6} l a s o c a n l a p Theo bai ra ta co: a^ + 32 + 33 +1 = 34 + 35 + 3g (1)
M a 3^ , 3 2 , 3 3 , 3 4 , 3 3 , 3 6 e {1,2,3,4,5,6} V3 doi mot khac nhau nen
a j + 3 2+ 3 3 + 3 4 + 3 5 + 3 ^ , = 1 + 2 + 3 + 4 + 5 + 6 = 21 (2) ^,.,4 T i r ( l ) , (2) suy r3: a, +32 +33 =10 „ <: ,, , 5; 0 ô,v'o:
Phuong t r i n h nsy c o e s e b p nghiem Is: (31,32,83) = (1,3,6); (1,4,5); (2,3,5) V o i moi bp ta CO 31.31 = 36 so'. ' > • , '
Vsy CO ea thay 3.36 = 108 so can lap. ^^•'^'''•'•^ V ' • '
* Cach 2: Gpi x = abcdef I3 so C3n Isp
a + b + e + d + e + f = l + 2 + 3 + 4 + 5 + 6 = 21 " ' ' - f : "
3+ b + c = d + e + f + l Ta co: <
= > 3 + b + c = l l . D o 3 , b , c e {1,2,3,4,5,6} -"'A'A h - f i n a Gffi! A-^s.; - ,.A Suy r 3 ta CO cac cap sau: (a,b,e) = (1,4,6); (2,3,6); (2,4,5) ' " - s ^ ^ ,A V a i m o i bp n h u vay ta CO 3! each chpn a,b,e va 3! eachchpn d,e,f D o do co: 3.31.3! = 108 so thoa man yeu cau bai toan.
V i d u 11. T u cac so 1,2,3 lap dupe bao nhieu so t u nhien g o m 6 chij' so thoa m a n dong thoi hai dieu kien sau
1. T r o n g m o i so, m o i chii so co mat d u n g mot Ian
2, Trong m o i so, hai c h u so giong nhau khong d u n g canh nhau. , G i a i
Dat A = (1,2,31 • Gpi S la tap cac so thoa m a n yeu cau t h u nhat ciia bai toan
6! . . Ta CO so cac so thoa man dieu k i ^ n t h i i nha't ciis bai toan la ^ = 90 (vi C3e so CO dang aabbee vs k h i h o 3 n vj hsi so 3,3 ta dupe so khong doi)
Gpi S j ,8 2 , 5 3 la tap cac so thupe S ma eo 1,2,3 cap chi> so giong nhau d u n g canh nhau. '
• So phan t u ciia S3 chinh bang so hoan v i ciia 3 cap 11,22,33 nen
• So phan t u ciia Sj chinh bSng so hoan vj ciia 4 phan tir la co dang 3,a,bb,cc n h u n g 3,3 khong d u n g csnh n h 3 u . Nen = ^ - 6 = 6 p h 3 n t u .
S, = 6
Phdn loai va phumtg phdp gidi Dai so- Gidi tich 11
• So p h a n t u cua S, c h i n h bang so hoan v i ci'ia cac phan t u c6 dang a,a,b,b,cc n h u n g a, a va b , b k h o n g d u n g canh nhau nen
Sj| = —- 6 - 1 2 = 1 2 .
Vay so cac s o t h o a m a n yeu cau bai toan la: 90 - (6 + 6 + 1 2 ) = 66 .
V i d u 12 H o i c6 tat ca bao nhieu so t u nhien chia he't cho 9 ma m o i so 2011 chi> so va t r o n g do c6 i t nha't hai chi> so 9 .'
G i a i
D a t X la cac so t u nhien thoa m a n yeu cau bai toan.
A = ( cac so t u n h i e n k h o n g v u a t qua 2011 c h u so va chia he't cho 9)
V o i m o i so thuoc A co m chir so ( m < 2008) t h i ta c6 the bo sung them 2011 - m so 0 vao phia t r u d c t h i so c6 d u g c k h o n g d o i k h i chia cho 9.
D o d o t a x e t c a c s o thuoc A c o d a n g a^aj-.-ajQi,; aj € |0,1,2,3,...,9 Ag = |a G A I ma t r o n g a k h o n g c6 c h u so 9|
A j = {a e A I ma t r o n g a c6 d i i n g 1 c h u so 9) j i f Ws!;.
g2on _ - | t../
• Ta thay tap A c6 1 + p h a n tir
fil'
• T i n h so p h a n tir cua A Q
V o i X e A n ^0 X = a , . . . a 2 o „ ; a i e{0, l,2,...,8} i = 1,2010 va a 2 o„ = 9- r vol 2010
r 6[ l; 9] , r = ^ aj . T u do ta suy ra A,, c6 9 ^ ' " " p h a n t u
i = i
• T i n h so p h a n tir cua A^ w ) 5 De lap so ciia thuoc tap A , ta thuc hien lien tiep hai b u o c sau:
B u a c 1: L a p m o t day g o m 2010 chu' so thuoc tap {0J,2...,8} va t o n g cac chir so chia he't cho 9. So cac day la 9^'"^^
B u a c 2: V o i m o i day vira lap tren, ta bo sung so 9 vao m o t vj t r i bat k i 6 day tren, ta c6 2010 each bo sung so 9
D o do A , CO 2010.9^"°^ phan tir.
Vay so cac so can lap la:
92011 _ i
8 - 2 0 1 0 . 9 ^ * " " + 8
C t y TNHH MTV DWH Khang Vict I I I . C A C B A I T O A N L U Y E N T A P v r n ; ; . ; : B a i l .
1. Ban can m u a m o t ao so m i ca 30 hoac 32. A o co 30 c6 3 m a u khac nhau, ao CO 32 CO 4 m a u khac n h a u . H o i ban co bao nhieu each l u a chon?
2. Co 10 c u o n sach T o a n khac nhau, 11 cuon sach V a n khac nhau va 7 cuon sach A n h van khac n h a u . M o t hoc sinh dug-c chon m g t q u y e n sach t r o n g cac
q u y e n sach tren. H o i co bao n h i e u each l u a chon. , + 3. Co bao n h i e u each xep 5 cuo'n sach Toan, 6 c u o n sach L y va 8 cuo'n sach
H o a len m o t ke sach sao cho cac c u o n sach c u n g m o t m o n hoc t h i xep canh nhau, bie't cac c u o n sach d o i m o t khac nhau .
G i a i
1. C o n g viec ta can t h u c hien t r o n g bai toan nay la m u a m o t chie'c ao so m i co 30 hoac 32. D e t h u c hien cong vi$c nay ta co hai p h u o n g an. & A M I (|> yCi Phuang tin 1: M u a ao co 30: P h u o n g an nay ta co 3 each chon (chon m o t t r o n g ba m a u ) .
Phuong an 2: M u a ao co 32: P h u o n g an nay ta co 4 each chpn. * o> *tC'^'^
Vay ta co ca thay 3 + 4 = 7 each lua chon. 'b'•n-no >•>; rhti mh noH"):, 2. De chon m o t cuon sach t r o n g nhii-ng cuo'n sach tren ta co cac p h u o n g an sau.
Phuong an 1: Cuo'n sach chon la cuon sach Toan: Ta co 10 each chon . . . Phuang an 2: C u o n sach chon la c u o n sach V a n : Ta co 11 each chon ! , j Phuong an 3: C u o n sach chon la c u o n sach A n h van: Ta co 7 each c h o n Vay CO 10 + 11 + 7 = 28 each l u a chon. ' , •.. 4 3. Ta xep cac cuo'n sach c i i n g m o t bo m o n t h a n h m o t n h o m
T r u o c he't ta xep 3 n h o m len ke sach chung ta co: 3! = 6 each xep
V o i m o i each xep 3 n h o m do len ke ta co 5! each hoan v i cac cuon sach Toan, 6! each hoan vj cac cuon sach L y va 8! each hoan vj cac cuon sach Hoa
Vay theo q u y tac n h a n co tat ca: 6.51.6!.8! each xep. f.i> i : te iib. nM i^M ' Bai 2. lif^i;.;- f\i f\(!ri ;jip,rt 0 o f b ' i o j j n oih.^Sr. iltj^js'n
1. Co bao n h i e u each xep 4 n g u o i A, B , C , D len 3 toa tau, biet m o i toa co the chua 4 n g u o i .
2. T r o n g m o t g i a i t h i da'u b o n g da co 20 d o i t h a m gia v o i the t h u c t h i da'u v o n g t r o n . C u hai d o i t h i gap n h a u d u n g m o t Ian. H o i co tat ca bao n h i e u tran da'u xay ra .
3- T i r t h a n h p h o A co 10 con d u o n g d i den thanh pho B, tir t h a n h pho A co 9 con d u o n g d i d e n t h a n h p h o C, tir B den D co 6 con d u o n g , tir C den D co 11 con d u o n g va k h o n g co con d u o n g nao n o i B v o i C. H o i co bao nhieu each d i tir A d e n D .
. . . a**
Phlin loiii i>ti jihuinig />/;<!/) ^iai Dai so — Giai tich 11
4. H o i dong quan trj ciia cong ty X gom 10 nguoi. H o i ccS bao nhieu each bau ra ba n g u o i vao ba v i t r i chvi tich, pho chu tich va t h u k i , bie't kha nang m o i n g u o i la n h u nhau.
' . ' • • - Giai ^ • •: y:^.,
1. De xep A ta C O 3 each len m o t trong ba toa -'^•••^•u P.<.',,^ i'^ ii'-i.,-!:Tir,u' i ,
Voi m o i each xep A ta c6 3 each xep B len toa tau > , , .. ,ô , Voi m o i each xep A,B ta c6 3 each xep C len toa tau . : .
Voi m o i each xep A,B,C ta eo 3 each xep D len toa tau ' Vay C O 3.3.3.3 = 81 each xep 4 nguoi len toa tau. ' ' ' " *' ' ' ''
2. C u m o i d o i phai thi dau v o i 19 doi eon lai nen eo 19.20 tran dau. T u y nhien theo each tinh nay thi mot tran dau chSng han A gap B dugc tinh hai Ian.
Do do so tran dau thuc tedien ra la: ^^"^^ = 190 tran.
2 ' nkf f.j 'iHjj;,,-5jjo.j 3. De d i t u A deh D ta c6 cae each d i sau ,FU^J:> ff^liiyi^if .' i n n ? ! 'If
A ^ B - ằ D : C 6 10.6 = 60 each •]{, },rp>'j!)'\, - H : , . v \u'/;;^'inwv<,'^'i A ^ . C ^ D : Co 9.11 = 99 each
Vay C O tat ca 159 each d i t u A den D .
4. Chon chu tjch c6 10 each chon, pho chu tjch c6 9 each va t h u k i c6 8 each.
Do do C O tat ca 10.9.8 = 720 each chon. vdo ^rnvi! W >fô riitih U'rin tnrh "iG . B a i S . . . . t • i . . j f c t i , - i : ni>.'|^m:ằôii!'l 1. Co 3 nam va 3 ni> can xep ngoi vao mot hang ghe! Hoi c6 may each xep sao cho:
a) N a m , nix ngoi xen ke?
b) N a m , nCr ngoi xen ke va c6 mot nguoi nam A, mot n g u o i ni> B phai ngoi ke nhau?
c) N a m , nCr ngoi xen ke va eo mot nguoi nam C, m o t nguoi n i i D khong dupe ngoi ke nhau?
2. M o t ban dai eo 2 day ghe doi d i f n nhau, m o i day g o m eo 6 ghe. N g u o i ta m u o n xep cho ngoi cho 6 hoe sinh truong A va 6 hoe sinh truong B vao ban noi tren. H o i c6 bao nhieu each xep cho ngoi trong m o i t r u o n g h o p sau:
a) Bat ki 2 hpe sinh nao ngoi canh nhau hoae doi dien nhau thi khac truong nhau.
b) Bat k i 2 hoc sinh nao ngoi doi dien nhau t h i khac truong nhau.
Giai
1. a) Co 6 each chon m o t nguoi tuy y ngoi vao cho t h u nha't. Tiep den, eo 3 each chon m g t nguoi khac phai ngoi vao ch6 t h u 2. Lai eo 2 each chon mot n g u o i khac phai ngoi vao cho t h u 3, c6 2 each chon vao cho t h u 4, c6 1 each chpn vao cho t h u 5, c6 1 each chon vao cho t h u 6.
V a y e o : 6.3.2.2.1.1 = 72 each.
11 Cty TNHH MTV DWH Khaug Vift
2.
b) Cho cap nam n u A, B do ngoi vao cho t h u nha't va cho t h u hai, c6 2 each.
Tiep den, cho t h u ba c6 2 each chon, cho t h u t u eo 2 each chon, ch6 t h u nam C O 1 each chon, cho t h u sau c6 1 each chpn.
Bay gio, cho cap nam n u A, B do ngoi vao cho t h u hai va cho t h u ba. K h i do, cho t h u nha't c6 2 each chon, cho t h u t u c6 2 each chon, cho t h u nam c6 1 each chon, cho t h u sau c6 1 each chon.
T u o n g t u k h i cap nam n u A, B do ngoi vao cho t h u ba va t h i i t u , t h u t u va t h u nam, t h u nam va t h u sau. ,^ ,, , V a y e o : 5.2.2.2.1.1. = 40 each.
c) So each chon de cap nam n u do khong ngoi ke nhau bang so each chon tuy y t r u so each chon de cap nam nu do ngoi ke nhau.
Vay C O : 7 2 - 4 0 = 32 each. . . , . . , Ta danh so lien tiep 12 cho ngoi bang cae so t u 1 den 6 thuoc m o t day va tir 7 den 12 thuoc m o t day
1 2 3 4 5 6 12 11 10
a)
V i tri 1 2 3 4 5 6 7 8 9 10 11 12
So each xep
12 6 5 5 4 4 3 3 2 2 1 1
V a y e o 12.6.5^.413^2^1 = 1036800 each xep.
b)
V i t r i 1 12 2 11 3 10 4 9 5 8 6 7
So each xep
12 6 10 5 8 4 6 3 4 2 2 1
Vay eo: 12.6^.10.5.8.42.3.22.1 = 33177600 each xep.
Bai 4.
1. Cho cae chCr so 1, 2, 3,..., 9. Tir cae so do c6 the lap dupe bao nhieu so?
a) Co 4 chCr so doi mot khac nhau.
b) So chSn gom 4 ehir so khac nhau va khong vupt qua 2011.
2. Co 100000 ve dupe danh so t u 00000 den 99999. H o i so ve g o m 5 c h u so khac nhau.
3. T i n h tong cae chir so gom 5 ehi> so dupe lap t u eac so 1, 2, 3, 4, 5?
• - " ' ,:-'nrằ'-5 '^A'-i: . ?i Giai I: r.^>u..- • •••
1. G p i s o c a n l a p x = abcd, a, b , c , d e {1,2,3,4,5,6,7,8,9' a) Co 9.8.7.6 = 3024 so
Phdn loai va phumtg phdp giai Dai so'-Gidi ttch 11
b) V i X clian nen d e {2,4,6,8} . Dong thai x < 2 0 1 1 = > a = l
• a = 1 => a CO 1 each chon, khi do d co 4 each chon; b,c c6 7.6 each
Suy r a c6: 1.4.6.7 = 168 so , ;
2. Gpi so in tren ve CO dang a , a 2 a 3 a 4 a c i U, : > \ (J ;
So'each chon a I la 10 ( a , cothelaO). i > v So each chon a j la 9.
So each chon a ^ la 8.
So each chon a ^ la 7. ' So each chon a ^ la 6.
Vay so ve gom 5 c h u so khac nhau : 10.9.8.7.6 = 30240.
3. Co 120 so CO 5 chu so duoe lap tir 5 chu so da cho. ' ^ ^ W B ,f Bay gia ta xet vj tri cua mot chu so trong 3 so 1, 2, 3, 4, 5 ch^ng han ta xet so 1.
So 1 CO the xe'p o 5 vi t r i khac nhau, moi vj tri co 4!=24 so nen khi ta nhom
; cac cac v i t r i nay lai co tong la : 24|l0''' + 10"* +10'' + 10^ + 10 + I j = 24.11111 Vay t o n g c a c s o c o S c h L r s o i a : 24.11111(1 + 2 + 3 + 4 + 5) = 5599944 . • Bai 5. T u cac so 1,2,3,4,5,6,7 lap du^c bao nhieu so t u nhien gom 4 chii so
khac nhau v a l a : ' ; ; " i • | " i l . S o c h a n * ( i : . v i - 2. Sole
3. So chia het cho 5 4. Tong hai chu so dau b^ng long hai chi> so cuoi.
G i a i
Goisocanlap x = abcd; a,b,c,d € {1,2,3,4,5,6,7} va a,b,c,d doi mot khacinhau.
1. Cong viec ta can thuc hien la lap so x thoa man x la so chin nen d phai la so chan. Do do de thuc hien cong viec nay ta thuc hien qua cac cong doan sau:
Bmc 1: Chon d : V i d la so chKn nen d chi co the la cac so 2,4,6 nen d co 3 each chon. M ' - ' ' . . . - , < . ' * , .
Bif&c 2: Chon a : V i ta da chon d nen a chi co the chon mot trong cac so cua tap {1,2,3,4,5,6,7} \l nen c6 6 each chon a
Bicac 3: Chon b : T u a n g t u ta co 5 each chon b Binrc 4: Chon c : Co 4 each chon.
Vay theo quy t3c nhan co: 3.6.5.4 = 360 so thoa man yeu cau bai toan.
2. V i so X can lap la so le nen d phai la so le. Ta lap x qua cac cong doan sau:
Bzrar 2; Co 4 each chon d Biror 2. Co 6 each chon a
Bi/'or3; C6 5 each chon b ezcor 4. CcS 4 each chon e j Vay CO 480 so thoa man yeu cau bai toan. ii,
' yi/lSIUM,.! . •
Cty TNHH MTV DWU Khana V't't 3 V i X chia het cho 5 nen d chi co the la 5 => co 1 each chon d.
Co 6 each chon a, 5 each chon b va 4 each chon c. ; , Vay c(S 1.6.5.4 = 120 so thoa man yeu cau bai toan. , Bai 6. Cho tap A = {1,2,3,4,5,6,7,8} • > 'Vi ,,u'V l \ , , ,