Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 119390 by mr W last updated on 24/Oct/20

we have 15 different mathematics books, 10 different physics books and 12 different chemistry books. we should choose 6 books such that they contain all three kinds of books. in how many ways can we do this?

$${we}\:{have}\:\mathrm{15}\:{different}\:{mathematics} \\ $$$${books},\:\mathrm{10}\:{different}\:{physics}\:{books}\:{and} \\ $$$$\mathrm{12}\:{different}\:{chemistry}\:{books}.\:{we}\:{should} \\ $$$${choose}\:\mathrm{6}\:{books}\:{such}\:{that}\:{they}\:{contain} \\ $$$${all}\:{three}\:{kinds}\:{of}\:{books}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{we}\:{do}\:{this}? \\ $$

Answered by bemath last updated on 24/Oct/20

(1) (4,1,1)⇒((P_4 ^(15) ×P_1 ^(10) ×P_1 ^(12) )/(3!)) (2)(3,2,1)⇒((P_3 ^(15) ×P_2 ^(10) ×P_1 ^(12) )/(3!)) (3)(3,1,2)⇒((P_3 ^(15) ×P_1 ^(10) ×P_2 ^(12) )/(3!)) (4)(2,3,1)⇒((P_2 ^(15) ×P_3 ^(10) ×P_1 ^(12) )/(3!)) (5)(2,2,2)⇒((P_2 ^(15) ×P_2 ^(10) ×P_2 ^(12) )/(3!)) (6)(2,1,3)⇒((P_2 ^(15) ×P_1 ^(10) ×P_3 ^(12) )/(3!)) (7)(1,4,1)⇒((P_1 ^(15) ×P_4 ^(10) ×P_1 ^(12) )/(3!)) (8)(1,3,2)⇒((P_1 ^(15) ×P_3 ^(10) ×P_2 ^(12) )/(3!)) (9)(1,2,3)⇒((P_1 ^(15) ×P_2 ^(10) ×P_3 ^(12) )/(3!)) (10)(1,1,4)⇒((P_1 ^(15) ×P_1 ^(10) ×P_4 ^(12) )/(3!)) Totally = (1)+(2)+(3)+...+(10)

$$\left(\mathrm{1}\right)\:\left(\mathrm{4},\mathrm{1},\mathrm{1}\right)\Rightarrow\frac{{P}_{\mathrm{4}} ^{\mathrm{15}} ×{P}_{\mathrm{1}} ^{\mathrm{10}} ×{P}_{\mathrm{1}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{2}\right)\left(\mathrm{3},\mathrm{2},\mathrm{1}\right)\Rightarrow\frac{{P}_{\mathrm{3}} ^{\mathrm{15}} ×{P}_{\mathrm{2}} ^{\mathrm{10}} ×{P}_{\mathrm{1}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{3}\right)\left(\mathrm{3},\mathrm{1},\mathrm{2}\right)\Rightarrow\frac{{P}_{\mathrm{3}} ^{\mathrm{15}} ×{P}_{\mathrm{1}} ^{\mathrm{10}} ×{P}_{\mathrm{2}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{4}\right)\left(\mathrm{2},\mathrm{3},\mathrm{1}\right)\Rightarrow\frac{{P}_{\mathrm{2}} ^{\mathrm{15}} ×{P}_{\mathrm{3}} ^{\mathrm{10}} ×{P}_{\mathrm{1}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{5}\right)\left(\mathrm{2},\mathrm{2},\mathrm{2}\right)\Rightarrow\frac{{P}_{\mathrm{2}} ^{\mathrm{15}} ×{P}_{\mathrm{2}} ^{\mathrm{10}} ×{P}_{\mathrm{2}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{6}\right)\left(\mathrm{2},\mathrm{1},\mathrm{3}\right)\Rightarrow\frac{{P}_{\mathrm{2}} ^{\mathrm{15}} ×{P}_{\mathrm{1}} ^{\mathrm{10}} ×{P}_{\mathrm{3}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{7}\right)\left(\mathrm{1},\mathrm{4},\mathrm{1}\right)\Rightarrow\frac{{P}_{\mathrm{1}} ^{\mathrm{15}} ×{P}_{\mathrm{4}} ^{\mathrm{10}} ×{P}_{\mathrm{1}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{8}\right)\left(\mathrm{1},\mathrm{3},\mathrm{2}\right)\Rightarrow\frac{{P}_{\mathrm{1}} ^{\mathrm{15}} ×{P}_{\mathrm{3}} ^{\mathrm{10}} ×{P}_{\mathrm{2}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{9}\right)\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\Rightarrow\frac{{P}_{\mathrm{1}} ^{\mathrm{15}} ×{P}_{\mathrm{2}} ^{\mathrm{10}} ×{P}_{\mathrm{3}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$$\left(\mathrm{10}\right)\left(\mathrm{1},\mathrm{1},\mathrm{4}\right)\Rightarrow\frac{{P}_{\mathrm{1}} ^{\mathrm{15}} ×{P}_{\mathrm{1}} ^{\mathrm{10}} ×{P}_{\mathrm{4}} ^{\mathrm{12}} }{\mathrm{3}!} \\ $$$${Totally}\:=\:\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)+...+\left(\mathrm{10}\right) \\ $$

Commented by mr W last updated on 24/Oct/20

why do you use P_k ^( n) , not C_k ^( n) ? when we choose 3 mathematics books, we don′t arrange them.

$${why}\:{do}\:{you}\:{use}\:{P}_{{k}} ^{\:{n}} ,\:{not}\:{C}_{{k}} ^{\:{n}} \:?\: \\ $$$${when}\:{we}\:{choose}\:\mathrm{3}\:{mathematics}\:{books}, \\ $$$${we}\:{don}'{t}\:{arrange}\:{them}. \\ $$

Commented by bemath last updated on 24/Oct/20

because the books are different

$${because}\:{the}\:{books}\:{are}\:{different} \\ $$

Commented by mr W last updated on 24/Oct/20

to choose 3 from 15 diff. mathematics books there are C_3 ^(15) ways, not P_3 ^(15)

$${to}\:{choose}\:\mathrm{3}\:{from}\:\mathrm{15}\:{diff}.\:{mathematics} \\ $$$${books}\:{there}\:{are}\:{C}_{\mathrm{3}} ^{\mathrm{15}} \:{ways},\:{not}\:{P}_{\mathrm{3}} ^{\mathrm{15}} \\ $$

Commented by bemath last updated on 24/Oct/20

sir the arrangement ABC not same to BAC

$${sir}\:{the}\:{arrangement}\:{ABC}\:{not}\:{same}\:{to}\:{BAC}\: \\ $$

Commented by mr W last updated on 24/Oct/20

we select 3 mathematics books A,B,C, don′t arrange them. so ABC,ACB, BAC etc. are the same.

$${we}\:{select}\:\mathrm{3}\:{mathematics}\:{books}\:{A},{B},{C}, \\ $$$${don}'{t}\:{arrange}\:{them}.\:{so}\:{ABC},{ACB}, \\ $$$${BAC}\:{etc}.\:{are}\:{the}\:{same}. \\ $$

Answered by ebi last updated on 24/Oct/20

let A=number of ways choosing 6 books from 3 kinds of books AND contain all 3 kinds of books. (maths,phy,chem) (1,1,4),(1,2,3),(1,3,2),(1,4,1), (2,1,3),(2,2,2),(2,3,1), (3,1,2),(3,2,1), (4,1,1) A =^(15) C_1 ^(10) C_1 ^(12) C_4 +^(15) C_1 ^(10) C_2 ^(12) C_3 + C_1 ^(10) C_3 ^(12) C_2 +^(15) C_1 ^(10) C_4 ^(12) C_1 + C_2 ^(10) C_1 ^(12) C_3 +^(15) C_2 ^(10) C_2 ^(12) C_2 + C_2 ^(10) C_3 ^(12) C_1 +^(15) C_3 ^(10) C_1 ^(12) C_2 + C_3 ^(10) C_2 ^(12) C_1 +^(15) C_4 ^(10) C_1 ^(12) C_1 =(15×10×495)+(15×45×220)+ (15×120×66)+(15×210×12)+ (105×10×220)+(105×45×66)+ (105×120×12)+(455×10×66)+ (455×45×12)+(1365×10×12) =74250+148500+118800+37800+ 231000+311850+151200+300300+ 245700+163800 =1783200

$$ \\ $$$${let}\:{A}={number}\:{of}\:{ways}\:{choosing}\:\mathrm{6}\:{books} \\ $$$${from}\:\mathrm{3}\:{kinds}\:{of}\:{books}\:{AND}\:{contain} \\ $$$${all}\:\mathrm{3}\:{kinds}\:{of}\:{books}. \\ $$$$ \\ $$$$\left({maths},{phy},{chem}\right) \\ $$$$\left(\mathrm{1},\mathrm{1},\mathrm{4}\right),\left(\mathrm{1},\mathrm{2},\mathrm{3}\right),\left(\mathrm{1},\mathrm{3},\mathrm{2}\right),\left(\mathrm{1},\mathrm{4},\mathrm{1}\right), \\ $$$$\left(\mathrm{2},\mathrm{1},\mathrm{3}\right),\left(\mathrm{2},\mathrm{2},\mathrm{2}\right),\left(\mathrm{2},\mathrm{3},\mathrm{1}\right), \\ $$$$\left(\mathrm{3},\mathrm{1},\mathrm{2}\right),\left(\mathrm{3},\mathrm{2},\mathrm{1}\right), \\ $$$$\left(\mathrm{4},\mathrm{1},\mathrm{1}\right) \\ $$$$ \\ $$$${A} \\ $$$$=^{\mathrm{15}} {C}_{\mathrm{1}} \:^{\mathrm{10}} {C}_{\mathrm{1}} \:^{\mathrm{12}} {C}_{\mathrm{4}} \:+\:^{\mathrm{15}} {C}_{\mathrm{1}} \:^{\mathrm{10}} {C}_{\mathrm{2}} \:^{\mathrm{12}} {C}_{\mathrm{3}} \:+ \\ $$$${C}_{\mathrm{1}} \:^{\mathrm{10}} {C}_{\mathrm{3}} \:^{\mathrm{12}} {C}_{\mathrm{2}} \:+\:^{\mathrm{15}} {C}_{\mathrm{1}} \:^{\mathrm{10}} {C}_{\mathrm{4}} \:^{\mathrm{12}} {C}_{\mathrm{1}} \:+ \\ $$$${C}_{\mathrm{2}} \:^{\mathrm{10}} {C}_{\mathrm{1}} \:^{\mathrm{12}} {C}_{\mathrm{3}} \:+\:^{\mathrm{15}} {C}_{\mathrm{2}} \:^{\mathrm{10}} {C}_{\mathrm{2}} \:^{\mathrm{12}} {C}_{\mathrm{2}} \:+ \\ $$$${C}_{\mathrm{2}} \:^{\mathrm{10}} {C}_{\mathrm{3}} \:^{\mathrm{12}} {C}_{\mathrm{1}} \:+\:^{\mathrm{15}} {C}_{\mathrm{3}} \:^{\mathrm{10}} {C}_{\mathrm{1}} \:^{\mathrm{12}} {C}_{\mathrm{2}} \:+ \\ $$$${C}_{\mathrm{3}} \:^{\mathrm{10}} {C}_{\mathrm{2}} \:^{\mathrm{12}} {C}_{\mathrm{1}} \:+\:^{\mathrm{15}} {C}_{\mathrm{4}} \:^{\mathrm{10}} {C}_{\mathrm{1}} \:^{\mathrm{12}} {C}_{\mathrm{1}} \\ $$$$=\left(\mathrm{15}×\mathrm{10}×\mathrm{495}\right)+\left(\mathrm{15}×\mathrm{45}×\mathrm{220}\right)+ \\ $$$$\:\:\:\:\left(\mathrm{15}×\mathrm{120}×\mathrm{66}\right)+\left(\mathrm{15}×\mathrm{210}×\mathrm{12}\right)+ \\ $$$$\:\:\:\:\left(\mathrm{105}×\mathrm{10}×\mathrm{220}\right)+\left(\mathrm{105}×\mathrm{45}×\mathrm{66}\right)+ \\ $$$$\:\:\:\:\left(\mathrm{105}×\mathrm{120}×\mathrm{12}\right)+\left(\mathrm{455}×\mathrm{10}×\mathrm{66}\right)+ \\ $$$$\:\:\:\:\left(\mathrm{455}×\mathrm{45}×\mathrm{12}\right)+\left(\mathrm{1365}×\mathrm{10}×\mathrm{12}\right) \\ $$$$=\mathrm{74250}+\mathrm{148500}+\mathrm{118800}+\mathrm{37800}+ \\ $$$$\:\:\:\:\mathrm{231000}+\mathrm{311850}+\mathrm{151200}+\mathrm{300300}+ \\ $$$$\:\:\:\:\mathrm{245700}+\mathrm{163800} \\ $$$$=\mathrm{1783200} \\ $$

Commented by mr W last updated on 24/Oct/20

correct answer, thanks!

$${correct}\:{answer},\:{thanks}! \\ $$

Commented by mr W last updated on 24/Oct/20

this is also the method i thought of. but if we have 4 or more kinds of books, this method is very tough. do you have any other ideas?

$${this}\:{is}\:{also}\:{the}\:{method}\:{i}\:{thought}\:{of}. \\ $$$${but}\:{if}\:{we}\:{have}\:\mathrm{4}\:{or}\:{more}\:{kinds}\:{of} \\ $$$${books},\:{this}\:{method}\:{is}\:{very}\:{tough}.\: \\ $$$${do}\:{you}\:{have}\:{any}\:{other}\:{ideas}? \\ $$

Commented by mr W last updated on 25/Oct/20

please try Q119515

$${please}\:{try}\:{Q}\mathrm{119515} \\ $$

Commented by ebi last updated on 25/Oct/20

i think solving probability problems are quite hard, as have to list all the possible outcomes and calculate them one by one. i still thinking to solve it in easier way. but, this method i think is the best way.

$${i}\:{think}\:{solving}\:{probability}\:{problems}\: \\ $$$${are}\:{quite}\:{hard},\:{as}\:{have}\:{to}\:{list}\:{all}\:{the}\: \\ $$$${possible}\:{outcomes}\:{and}\:{calculate}\:{them}\: \\ $$$${one}\:{by}\:{one}.\:{i}\:{still}\:{thinking}\:{to} \\ $$$${solve}\:{it}\:{in}\:{easier}\:{way}.\:{but},\:{this}\:{method}\: \\ $$$${i}\:{think}\:{is}\:{the}\:{best}\:{way}. \\ $$

Commented by mr W last updated on 25/Oct/20

thanks!

$${thanks}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com