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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!
$${thanks}! \\ $$

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