Hoan Vi Chinh Hop to Hop

ThS

CHUYN :HON V CHNH HP T HP

A. TM TT GIO KHOA V PHNG PHP GII TON

1. Hon v

nh ngha

Cho tp hp X gm n phn t phn bit . Mi cch sp xp n phn t ca X theo mt th t no c gi l mt hon v ca n phn t. S cc hon v ca n phn t c k hiu l Pn.

. Quy c: 0! = 1.

V d 1. Sp xp 5 ngi vo mt bng gh c 5 ch. Hi c bao nhiu cch.

Gii

Mi cch i ch 1 trong 5 ngi trn bng gh l 1 hon v.

Vy c P5 = 5! = 120 cch sp.

V d 2. T cc ch s 0, 1, 2, 3, 4 c th lp c my s t nhin c 5 ch s khc nhau.

Gii

Gi vi v phn bit l s cn lp.

+ Bc 1: ch s nn c 4 cch chn a1.

+ Bc 2: sp 4 ch s cn li vo 4 v tr c 4! = 24 cch.

Vy c 4.24 = 96 s.

2. Chnh hp

nh ngha

Cho tp hp X gm n phn t phn bit . Mi cch chn ra k phn t ca X v sp xp theo mt th t no c gi l mt chnh hp chp k ca n phn t. S cc chnh hp chp k ca n phn t c k hiu l .

.

Nhn xt:

.

V d 3. Sp xp 5 ngi vo mt bng gh c 7 ch. Hi c bao nhiu cch.

Gii

Mi cch chn ra 5 ch ngi t bng gh sp 5 ngi vo v c hon v l mt chnh hp chp 5 ca 7.

Vy c cch sp.

V d 4. T tp hp c th lp c my s t nhin c 4 ch s khc nhau.

Gii

Gi vi v phn bit l s cn lp.

+ Bc 1: ch s nn c 5 cch chn a1.

+ Bc 2: chn 3 trong 5 ch s cn li sp vo 3 v tr cch.

Vy c s.

3. T hp

nh ngha

Cho tp hp X gm n phn t phn bit . Mi cch chn ra k phn t ca X c gi l mt t hp chp k ca n phn t. S cc t hp chp k ca n phn t c k hiu l .

.

V d 5. C 10 cun sch ton khc nhau. Chn ra 4 cun, hi c bao nhiu cch.

Gii

Mi cch chn ra 4 trong 10 cun sch l mt t hp chp 4 ca 10.

Vy c cch chn.

V d 6. Mt nhm c 5 nam v 3 n. Chn ra 3 ngi sao cho trong c t nht 1 n. Hi c bao nhiu cch.

Gii

+ Trng hp 1: chn 1 n v 2 nam.

- Bc 1: chn ra 1 trong 3 n c 3 cch.

- Bc 2: chn ra 2 trong 5 nam c .

Suy ra c cch chn.


- Bc 1: chn ra 2 trong 3 n c cch.

- Bc 2: chn ra 1 trong 5 nam c 5.

Suy ra c cch chn.

+ Trng hp 3: chn 3 n c 1 cch.

Vy c cch chn.

V d 7. Hi c th lp c bao nhiu s t nhin c 4 ch s sao cho trong mi s , ch s hng ngn ln hn hng trm, ch s hng trm ln hn hng chc v ch s hng chc ln hn hng n v.

Gii

Gi vi l s cn lp.

.

T 10 phn t ca X ta chn ra 4 phn t bt k th ch lp c 1 s A. Ngha l khng c hon v hay l mt t hp chp 4 ca 10.

Vy c s.

Nhn xt:

i) iu kin xy ra hon v, chnh hp v t hp l n phn t phi phn bit.

ii) Chnh hp v t hp khc nhau ch l sau khi chn ra k trong n phn t th chnh hp c sp th t cn t hp th khng.

4. Phng php gii ton

4.1. Phng php 1

Bc 1. c k cc yu cu v s liu ca bi. Phn bi ton ra cc trng hp, trong mi trng hp li phn thnh cc giai on.

Bc 2. Ty tng giai on c th v gi thit bi ton s dng quy tc cng, nhn, hon v, chnh hp hay t hp.

Bc 3. p n l tng kt qu ca cc trng hp trn.

V d 8. Mt nhm cng nhn gm 15 nam v 5 n. Ngi ta mun chn t nhm ra 5 ngi lp thnh mt t cng tc sao cho phi c 1 t trng nam, 1 t ph nam v c t nht 1 n. Hi c bao nhiu cch lp t cng tc.

Gii


- Bc 1: chn 1 trong 5 n c 5 cch.

- Bc 2: chn 2 trong 15 nam lm t trng v t ph c cch.

- Bc 3: chn 2 trong 13 nam cn li c cch.

Suy ra c cch chn cho trng hp 1.


- Bc 1: chn 2 trong 5 n c cch.


- Bc 3: chn 1 trong 13 nam cn li c 13 cch.



- Bc 1: chn 3 trong 5 n c cch.



Vy c cch.

Cch khc:

+ Bc 1: chn 2 trong 15 nam lm t trng v t ph c cch.

+ Bc 2: chn 3 t vin, trong c n.

- Trng hp 1: chn 1 n v 2 nam c cch.

- Trng hp 2: chn 2 n v 1 nam c cch.

- Trng hp 3: chn 3 n c cch.

Vy c cch.

4.2. Phng php 2.

i vi nhiu bi ton, phng php 1 rt di. Do ta s dng phng php loi tr (phn b) theo php ton .

Bc 1. Chia yu cu ca thnh 2 phn l yu cu chung X (tng qut) gi l loi 1 v yu cu ring A. Xt l ph nh ca A, ngha l khng tha yu cu ring gi l loi 2.

Bc 2. Tnh s cch chn loi 1 v loi 2.

Bc 3. p n l s cch chn loi 1 tr s cch chn loi 2.

Ch :

Cch phn loi 1 v loi 2 c tnh tng i, ph thuc vo ch quan ca ngi gii.

V d 9. T cc ch s 0, 1, 2, 3, 4 c th lp c my s t nhin c 5 ch s khc nhau.

Gii

+ Loi 1: ch s a1 ty , ta c 5! = 120 s.

+ Loi 2: ch s a1 = 0, ta c 4! = 24 s.

Vy c 120 24 = 96 s.

V d 10. Mt nhm c 7 nam v 6 n. Chn ra 3 ngi sao cho trong c t nht 1 n. Hi c bao nhiu cch.

Gii

+ Loi 1: chn 3 ngi ty trong 13 ngi c cch.

+ Loi 2: chn 3 nam (khng c n) trong 7 nam c cch.

Vy c cch chn.

V d 11. T 20 cu hi trc nghim gm 9 cu d, 7 cu trung bnh v 4 cu kh ngi ta chn ra 10 cu lm kim tra sao cho phi c c 3 loi d, trung bnh v kh. Hi c th lp c bao nhiu kim tra.

Gii

+ Loi 1: chn 10 cu ty trong 20 cu c cch.

+ Loi 2: chn 10 cu c khng qu 2 trong 3 loi d, trung bnh v kh.

- Trng hp 1: chn 10 cu d v trung bnh trong 16 cu c cch.

- Trng hp 2: chn 10 cu d v kh trong 13 cu c cch.

- Trng hp 3: chn 10 cu trung bnh v kh trong 11 cu c cch.

Vy c kim tra.

Ch :

Gii bng phng php phn b c u im l ngn tuy nhin nhc im l thng sai st khi tnh s lng tng loi.

V d 12. T 20 cu hi trc nghim gm 9 cu d, 7 cu trung bnh v 4 cu kh ngi ta chn ra 7 cu lm kim tra sao cho phi c c 3 loi d, trung bnh v kh. Hi c th lp c bao nhiu kim tra.

Cch gii sai:


+ Loi 2: chn 7 cu khng tha yu cu.

- Trng hp 1: chn 7 cu d trong 9 cu c cch.

- Trng hp 2: chn 7 cu trung bnh c 1 cch.

- Trng hp 3: chn 7 cu d v trung bnh trong 16 cu c cch.



Vy c kim tra!

Sai st trong cch tnh s loi 2. Chng hn, khi tnh s trong trng hp 3 ta tnh lp li trng hp 1 v trng hp 2.

Cch gii sai khc:



- Trng hp 1: chn 7 cu d hoc trung bnh trong 16 cu c cch.

- Trng hp 2: chn 7 cu d hoc kh trong 13 cu c cch.

- Trng hp 3: chn 7 cu trung bnh hoc kh trong 11 cu c cch.

Vy c kim tra.

Sai st do ta tnh lp li s cch chn ch c 7 cu d v ch c 7 cu trung bnh trong trng hp 1 v trng hp 2.

Cch gii ng:



- Trng hp 1: chn 7 cu d hoc trung bnh trong 16 cu c cch.



Vy c kim tra.

V d 13. Hi ng qun tr ca mt cng ty gm 12 ngi, trong c 5 n. T hi ng qun tr ngi ta bu ra 1 ch tch hi ng qun tr, 1 ph ch tch hi ng qun tr v 2 y vin. Hi c my cch bu sao cho trong 4 ngi c bu phi c n.

Gii

+ Loi 1: bu 4 ngi ty (khng phn bit nam, n).

- Bc 1: bu ch tch v ph ch tch c cch.

- Bc 2: bu 2 y vin c cch.

Suy ra c cch bu loi 1.

+ Loi 2: bu 4 ngi ton nam.

- Bc 1: bu ch tch v ph ch tch c cch.

- Bc 2: bu 2 y vin c cch.

Suy ra c cch bu loi 2.

Vy c cch.

5. Hon v lp (tham kho)

Cho tp hp X c n phn t gm n1 phn t ging nhau, n2 phn t khc li ging nhau, , nk phn t khc na li ging nhau . Mi cch sp n phn t ny vo n v tr l mt hon v lp, s hon v lp l .

V d 14. T cc ch s 1, 2, 3 lp c bao nhiu s t nhin c ng 5 ch s 1, 2 ch s 2 v 3 ch s 3.

Gii

Xem s cn lp c 10 ch s gm 5 ch s 1 ging nhau, 2 ch s 2 ging nhau v 3 ch s 3 ging nhau.

Vy c s.

Cch gii thng dng:

+ Bc 1: chn 5 trong 10 v tr sp 5 ch s 1 c cch.

+ Bc 2: chn 2 trong 5 v tr cn li sp 2 ch s 2 c cch.

+ Bc 3: sp 3 ch s 3 vo 3 v tr cn li c 1 cch.

Vy c s.

B. BI TP

Bi 1. Cn xp 3 nam v 2 n vo 1 hng gh c 7 ch ngi sao cho 3 nam ngi k nhau v 2 n ngi k nhau. Hi c bao nhiu cch.

Bi 2. Xt a gic u c n cnh, bit s ng cho gp i s cnh. Tnh s cnh ca a gic u .

Bi 3. Tnh s cc s t nhin i mt khc nhau c 6 ch s to thnh t cc ch s 0, 1, 2, 3, 4, 5 sao cho 2 ch s 3 v 4 ng cnh nhau.

Bi 4. Tnh s cc s t nhin c 4 ch s i mt khc nhau c thnh lp t 0, 1, 2, 3, 4, 5 sao cho trong mi s u c mt t nht ch s 1 hoc 2.

Bi 5. Hai nhm ngi cn mua nn nh, nhm th nht c 2 ngi v h mun mua 2 nn k nhau, nhm th hai c 3 ngi v h mun mua 3 nn k nhau. H tm c mt l t chia thnh 7 nn ang rao bn (cc nn nh nhau v cha c ngi mua). Tnh s cch chn nn ca mi ngi tha yu cu trn.

Bi 6. T 4 ch s 0, 1, 2, 3 lp thnh cc s t nhin c 3 ch s phn bit. Tnh tng cc s c thnh lp.

Bi 7. Tnh s hnh ch nht c to thnh t 4 trong 20 nh ca a gic u c 20 cnh ni tip ng trn tm O.

Bi 8. Cho a gic u c 2n cnh ni tip ng trn tm O. Bit s tam gic c cc nh l 3 trong 2n nh ca a gic nhiu gp 20 ln s hnh ch nht c cc nh l 4 trong 2n nh ca a gic. Tnh s hnh ch nht.

Bi 9. i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 em khi 12, 6 em khi 11 v 5 em khi 10. Tnh s cch chn 6 em trong i i d tri h sao cho mi khi c t nht 1 em c chn.

Bi 10. Cho tp hp X gm 10 phn t khc nhau. Tnh s tp hp con khc rng cha mt s chn cc phn t ca X.

Bi 11. Mt hp ng 15 vin bi khc nhau gm 4 bi , 5 bi trng v 6 bi vng. Tnh s cch chn 4 vin bi t hp sao cho khng c 3 mu.

Bi 12. Gii v ch bng Quc gia c 14 i tham gia thi u vng trn 1 lt, bit rng trong 1 trn u: i thng c 3 im, ha 1 im, thua 0 im v c 23 trn ha. Tnh s im trung bnh ca 1 trn trong ton gii.

Bi 13. Tnh s cc s t nhin gm 7 ch s c chn t 1, 2, 3, 4, 5 sao cho ch s 2 c mt ng 2 ln, ch s 3 c mt ng 3 ln v cc ch s cn li c mt khng qu 1 ln.

Bi 14. Tnh s cc s t nhin gm 5 ch s phn bit v mt trong 3 ch s u tin l 1 c thnh lp t cc ch s 0, 1, 2, 3, 4, 5, 6, 7.

Bi 15. T mt nhm 30 hc sinh gm 15 hc sinh khi A, 10 hc sinh khi B v 5 hc sinh khi C chn ra 15 hc sinh sao cho c t nht 5 hc sinh khi A v c ng 2 hc sinh khi C. Tnh s cch chn.

Bi 16. T mt nhm 12 hc sinh gm 4 hc sinh khi A, 4 hc sinh khi B v 4 hc sinh khi C chn ra 5 hc sinh sao cho mi khi c t nht 1 hc sinh. Tnh s cch chn.

Bi 17. Tnh s tp hp con ca X = {0; 1; 2; 3; 4; 5; 6} cha 1 m khng cha 0.

Bi 18. i thanh nin xung kch ca mt trng ph thng c 12 hc sinh gm 5 hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Tnh s cch chn 4 hc sinh i lm nhim v sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn.

Bi 19. T cc ch s 0, 1, 2, 3, 4, 5, 6 lp thnh s t nhin chn c 5 ch s phn bit nh hn 25000. Tnh s cc s lp c.

Bi 20. Tp hp A gm n phn t (n 4). Bit rng s tp hp con cha 4 phn t ca A bng 20 ln s tp hp con cha 2 phn t ca A, tm s sao cho s tp hp con cha k phn t ca A l ln nht.

C. HNG DN GII

Bi 1. Xt 3 loi gh gm 1 gh c 3 ch, 1 gh c 2 ch v 2 gh c 1 ch ngi.

+ Bc 1: do 2 gh c 1 ch khng phn bit nn chn 2 trong 4 v tr sp gh 2 v 3 ch ngi c cch.

+ Bc 2: sp 3 nam vo gh 3 ch c 3! = 6 cch.

+ Bc 3: sp 2 n vo gh 2 ch c 2! = 2 cch.

Vy c 12.6.2 = 144 cch sp.

Bi 2. Chn 2 trong n nh ca a gic ta lp c 1 cnh hoc ng cho.

S cnh v ng cho l . Suy ra s ng cho l .

Ta c:

.

Vy c 7 cnh.

Bi 3. Xt s c 5 ch s gm 0, 1, 2, 5 v ch s kp l (3, 4).

+ Loi 1: ch s hng trm ngn c th l 0.

- Bc 1: sp 5 ch s vo 5 v tr c 5! = 120 cch.

- Bc 2: vi mi cch sp ch s kp c 2 hon v ch s 3 v 4.

Suy ra c 120.2 = 240 s.

+ Loi 2: ch s hng trm ngn l 0.

- Bc 1: sp 4 ch s vo 4 v tr cn li c 4! = 24 cch.

- Bc 2: vi mi cch sp ch s kp c 2 hon v ch s 3 v 4.

Suy ra c 24.2 = 48 s.Vy c 240 48 = 192 s.

Bi 4.

+ Loi 1: ch s a1 c th l 0.

Sp 4 trong 6 ch s vo 4 v tr c cch. Sp 4 ch s 0, 3, 4, 5 vo 4 v tr c 4! = 24 cch. Suy ra c 360 24 = 336 s.

+ Loi 2: ch s a1 l 0 (v tr a1 c ch s 0).

Sp 3 trong 5 ch s vo 3 v tr c cch. Sp 3 ch s 3, 4, 5 vo 3 v tr c 3! = 6 cch. Suy ra c 60 6 = 54 s.

Vy c 336 54 = 282 s.

Cch khc:

+ Loi 1: S t nhin c 4 ch s ty .

- Bc 1: Chn 1 trong 5 ch s khc 0 sp vo a1 c 5 cch.

- Bc 2: Chn 3 trong 5 ch s khc a1 sp vo 3 v tr cn li c cch.

Suy ra c 5.60 = 300 s.

+ Loi 2: S t nhin c 4 ch s gm 0, 3, 4, 5 (khng c 1 v 2).

- Bc 1: Chn 1 trong 3 ch s khc 0 sp vo a1 c 3 cch.

- Bc 2: Sp 3 ch s cn li vo 3 v tr 3! = 6 cch.

Suy ra c 3.6 = 18 s.

Vy c 300 18 = 282 s.

Bi 5. Xem l t c 4 v tr gm 2 v tr 1 nn, 1 v tr 2 nn v 1 v tr 3 nn.

+ Bc 1: nhm th nht chn 1 v tr cho 2 nn c 4 cch v mi cch c 2! = 2 cch chn nn cho mi ngi. Suy ra c 4.2 = 8 cch chn nn.

+ Bc 2: nhm th hai chn 1 trong 3 v tr cn li cho 3 nn c 3 cch v mi cch c 3! = 6 cch chn nn cho mi ngi. Suy ra c 3.6 = 18 cch chn nn.

Vy c 8.18 = 144 cch chn nn cho mi ngi.

Bi 6. + Xt s A c 3 ch s phn bit v ch s hng trm c th l 0.

T s A ta lp c 12 cp s c tng l 333. V d 012 + 321 = 333.

Suy ra tng cc s A l 12.333 = 3996.

+ Xt s B c 3 ch s phn bit v ch s hng trm l 0.

T s B ta lp c 3 cp s c tng l 44. V d 032 + 012 = 44.

Suy ra tng cc s B l 3.44 = 132.

Vy tng cc s tha yu cu l 3996 132 = 3864.

Cch khc:

+ Xt s A c 3 ch s phn bit v ch s hng trm c th l 0.

- S cc s A l s. S ln cc ch s c mt hng trm, hng chc v n v l nh nhau v bng 24 : 4 = 6 ln.

- Tng cc ch s hng trm (hng chc, n v) ca 24 s l:

6.(0 + 1 + 2 + 3) = 36.

Suy ra tng cc s A l 36.(100 + 10 + 1) = 3996.

+ Xt s B c 3 ch s phn bit v ch s hng trm l 0.

- S cc s B l s. S ln cc ch s 1, 2, 3 c mt hng chc v n v l nh nhau v bng 6 : 3 = 2 ln.

- Tng cc ch s hng chc (n v) ca 6 s l 2.(1 + 2 + 3) = 12.

Suy ra tng cc s B l 12.(10 + 1) = 132.

Vy tng cc s tha yu cu l 3996 132 = 3864.

Bi 7. Nhn thy cc hnh ch nht c to thnh c 2 ng cho l ng knh ca ng trn. V ng thng d qua tm O v khng qua nh ca a gic u th d chia a gic thnh 2 phn, mi phn c 10 nh. Suy ra s ng cho ca a gic i qua tm O l 10. Chn 2 trong 10 ng cho th lp c 1 hnh ch nht.

Vy c hnh ch nht.

Bi 8. + L lun tng t cu 65 ta c hnh ch nht.

+ S tam gic to thnh t 3 trong 2n nh ca a gic l .

+ T gi thit ta c:

.

Vy c hnh ch nht.

Bi 9. Cch gii sai:

+ Chn ty 6 em trong i c cch.

+ Chn 6 em trong i thuc khi 12 hoc khi 11 c cch.



Vy c 18564 1716 924 462 = 15462 cch chn!

Sai ch lp 12 v lp 11 ta tnh lp li.

Cch gii ng:

+ Chn ty 6 em trong i c cch.


+ Chn 6 em trong i thuc khi 12 v khi 10 c cch.

+ Chn 6 em trong i thuc khi 11 v khi 10 c cch.

Vy c 18564 1716 917 461 = 15454 cch chn.

Bi 10. + S tp hp con cha 2 phn t ca X l .

+ S tp hp con cha 4 phn t ca X l .



+ S tp hp con cha 10 phn t ca X l 1.

Vy c 45 + 210 + 210 + 45 + 1 = 511 tp hp.

Bi 11. + Trng hp 1: chn 4 bi hoc trng c cch.

+ Trng hp 2: chn 4 bi v vng hoc 4 bi vng c cch.

+ Trng hp 3: chn 4 bi trng v vng c cch.

Vy c 126 + 209 + 310 = 645 cch.

Cch khc:

+ Loi 1: chn ty 4 trong 15 vin bi c cch.

+ Loi 2: chn c 3 mu c 720 cch gm cc trng hp sau:

- Chn 2 bi , 1 bi trng v 1 bi vng c 180 cch.



Vy c 1365 720 = 645 cch.

Bi 12. + Do thi u vng trn 1 lt nn 2 i bt k ch u vi nhau ng 1 trn. S trn u ca gii l .

+ Tng s im ca 2 i trong 1 trn ha l 2 nn tng s im ca 23 trn ha l 2.23 = 46.

+ Tng s im ca 2 i trong 1 trn khng ha l 3 nn tng s im ca 68 trn khng ha l 3.68 = 204.

Vy s im trung bnh ca 1 trn l im.

Bi 13. Xem s c 7 ch s nh 7 v tr thng hng.

+ Bc 1: chn 2 trong 7 v tr sp 2 ch s 2 (khng hon v) c cch.

+ Bc 2: chn 3 trong 5 v tr cn li sp 3 ch s 3 (khng hon v) c cch.

+ Bc 3: chn 2 trong 3 ch s 1, 4, 5 sp vo 2 v tr cn li (c hon v) c cch.

Vy c 21.10.6 = 1260 s.

Bi 14. + Loi 1: ch s a1 c th l 0.

- Bc 1: chn 1 trong 3 v tr u sp ch s 1 c 3 cch.

- Bc 2: chn 4 trong 7 ch s (tr ch s 1) sp vo cc v tr cn li c cch. Suy ra c 3.840 = 2520 s.

+ Loi 2: ch s a1 l 0.

- Bc 1: chn 1 trong 2 v tr th 2 v 3 sp ch s 1 c 2 cch.

- Bc 2: chn 3 trong 6 ch s (tr 0 v 1) sp vo cc v tr cn li c cch. Suy ra c 2.120 = 240 s.

Vy c 2520 240 = 2280 s.

Bi 15. + Loi 1: Chn 2 hc sinh khi C, 13 hc sinh khi B hoc khi A c cch.

+ Loi 2: Chn 2 hc sinh khi C, 13 hc sinh khi B v khi A khng tha yu cu.

- Trng hp 1: Chn 2 hc sinh khi C, 10 hc sinh khi B v 3 hc sinh khi A c cch.

- Trng hp 2: Chn 2 hc sinh khi C, 9 hc sinh khi B v 4 hc sinh khi A c cch.

Vy c cch.

Bi 16. + Trng hp 1: 1 khi c 3 hc sinh v 2 khi cn li mi khi c 1 hc sinh.

- Bc 1: chn 1 khi c 3 hc sinh c 3 cch.

- Bc 2: trong khi chn ta chn 3 hc sinh c cch.

- Bc 3: 2 khi cn li mi khi c 4 cch chn.

Suy ra c 3.4.4.4 = 192 cch.

+ Trng hp 2: 2 khi c 2 hc sinh v khi cn li c 1 hc sinh.

- Bc 1: chn 2 khi c 2 hc sinh c cch.

- Bc 2: trong 2 khi chn ta chn 2 hc sinh c cch.

- Bc 3: khi cn li c 4 cch chn.

Suy ra c 3.6.6.4 = 432 cch.

Vy c 192 + 432 = 624 cch.

Cch khc:

+ Chn 5 hc sinh ty c cch.

+ Chn 5 hc sinh khi A v B (tng t khi A v C, B v C) c cch.

Vy c 792 3.56 = 624 cch.

Bi 17. + S tp hp con khng cha phn t no ca l .

+ S tp hp con cha 1 phn t ca l .





Suy ra s tp hp con ca l . Ta hp cc tp hp con ny vi {1} th c 32 tp hp tha bi ton.

Bi 18. Cch gii sai:

+ Trng hp 1: chn 4 hc sinh lp A hoc lp B c cch.

+ Trng hp 2: chn 4 hc sinh lp A hoc lp C c cch.

+ Trng hp 3: chn 4 hc sinh lp B hoc lp C c cch.

Vy c cch!

Sai do ta tnh lp li trng hp ch chn 4 hc sinh lp A v trng hp ch chn 4 hc sinh lp B.

Cch gii sai khc:

+ Loi 1: chn ty 4 trong 12 hc sinh c cch.

+ Loi 2: chn 4 hc sinh c mt c 3 lp.

- Bc 1: chn 1 hc sinh lp A, 1 hc sinh lp B v 1 hc sinh lp C c:

5.4.3 = 60 cch.

- Bc 2: chn 1 hc sinh trong 9 hc sinh cn li ca 3 lp c 9 cch.

Suy ra c 9.60 = 540 cch chn loi 2 (ln hn s cch chn loi 1!).

Sai l do khi thc hin bc 1 v bc 2, v tnh ta to ra th t trong cch chn. C ngha l t t hp chuyn sang chnh hp!

Cch gii ng:

+ Loi 1: chn ty 4 trong 12 hc sinh c cch.

+ Loi 2: chn 4 hc sinh c mt c 3 lp, ta c 3 trng hp sau:

- Chn 2 hc sinh lp A, 1 hc sinh lp B v 1 hc sinh lp C c cch.



Vy c 495 (120 + 90 + 60) = 225 cch.

Bi 19. Gi s cn lp l vi .

+ Trng hp 1: a1 = 1.

C 4 cch chn a5 v cch chn cc ch s cn li nn c s.

+ Trng hp 2: a1 = 2, a2 l.

C 2 cch chn a2, 3 cch chn a5 v cch chn cc ch s cn li nn c s.

+ Trng hp 3: a1 = 2, a2 chn.

C 2 cch chn a2, 2 cch chn a5 v cch chn cc ch s cn li nn c s.

Vy c 240 + 72 + 48 = 360 s.

Bi 20. S tp hp con cha k phn t ca A l . Ta c:

.

Vy k = 9.21

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