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Question Number 124878 by mr W last updated on 06/Dec/20

20 students should stand in 5 different rows. each row should have at least 2 students. in how many ways can you arrange them?

$$\mathrm{20}\:{students}\:{should}\:{stand}\:{in}\:\mathrm{5} \\ $$$${different}\:{rows}.\:{each}\:{row}\:{should}\:{have} \\ $$$${at}\:{least}\:\mathrm{2}\:{students}.\:{in}\:{how}\:{many}\:{ways} \\ $$$${can}\:{you}\:{arrange}\:{them}? \\ $$

Answered by liberty last updated on 07/Dec/20

(1) [ −_2 ^R_1 , −_2 ^R_2 , −_2 ^R_3 , −_2 ^R_4 , −_(12) ^R_5 ] ⇒ ((5!)/(4!))×P_2 ^( 20) ×P_2 ^( 18) ×P_2 ^( 16) ×P_2 ^( 14) ×12! (2) [ (R_1 /2),(R_2 /2),(R_3 /2),(R_4 /3),(R_5 /(11)) ] ⇒((5!)/(3!))×P_2 ^( 20) ×P_2 ^( 18) ×P_2 ^( 16) ×P_3 ^( 14) ×11! (3) [ (R_1 /2),(R_2 /2),(R_3 /2),(R_4 /4),(R_5 /(10)) ] ⇒ ((5!)/(3!))×_(20) P_2 ×_(18) P_2 ×_(16) P_2 ×_(14) P_4 ×10! next...

$$\left(\mathrm{1}\right)\:\left[\:\underset{\mathrm{2}} {\overset{{R}_{\mathrm{1}} } {−}},\:\underset{\mathrm{2}} {\overset{{R}_{\mathrm{2}} } {−}}\:,\:\underset{\mathrm{2}} {\overset{{R}_{\mathrm{3}} } {−}}\:,\:\underset{\mathrm{2}} {\overset{{R}_{\mathrm{4}} } {−}}\:,\:\underset{\mathrm{12}} {\overset{{R}_{\mathrm{5}} } {−}}\:\right]\:\Rightarrow\:\frac{\mathrm{5}!}{\mathrm{4}!}×{P}_{\mathrm{2}} ^{\:\mathrm{20}} ×{P}_{\mathrm{2}} ^{\:\mathrm{18}} ×{P}_{\mathrm{2}} ^{\:\mathrm{16}} ×{P}_{\mathrm{2}} ^{\:\mathrm{14}} ×\mathrm{12}!\: \\ $$$$\left(\mathrm{2}\right)\:\left[\:\frac{{R}_{\mathrm{1}} }{\mathrm{2}},\frac{{R}_{\mathrm{2}} }{\mathrm{2}},\frac{{R}_{\mathrm{3}} }{\mathrm{2}},\frac{{R}_{\mathrm{4}} }{\mathrm{3}},\frac{{R}_{\mathrm{5}} }{\mathrm{11}}\:\right]\:\Rightarrow\frac{\mathrm{5}!}{\mathrm{3}!}×{P}_{\mathrm{2}} ^{\:\mathrm{20}} ×{P}_{\mathrm{2}} ^{\:\mathrm{18}} ×{P}_{\mathrm{2}} ^{\:\mathrm{16}} ×{P}_{\mathrm{3}} ^{\:\mathrm{14}} ×\mathrm{11}! \\ $$$$\left(\mathrm{3}\right)\:\left[\:\frac{{R}_{\mathrm{1}} }{\mathrm{2}},\frac{{R}_{\mathrm{2}} }{\mathrm{2}},\frac{{R}_{\mathrm{3}} }{\mathrm{2}},\frac{{R}_{\mathrm{4}} }{\mathrm{4}},\frac{{R}_{\mathrm{5}} }{\mathrm{10}}\:\right]\:\Rightarrow\:\frac{\mathrm{5}!}{\mathrm{3}!}×_{\mathrm{20}} {P}_{\mathrm{2}} ×_{\mathrm{18}} {P}_{\mathrm{2}} ×_{\mathrm{16}} {P}_{\mathrm{2}} ×_{\mathrm{14}} {P}_{\mathrm{4}} ×\mathrm{10}! \\ $$$${next}... \\ $$

Commented by mr W last updated on 07/Dec/20

thanks for trying! since we have 1001 cases, i don′t think we can solve in this way. but i know no other ways how to solve.

$${thanks}\:{for}\:{trying}! \\ $$$${since}\:{we}\:{have}\:\mathrm{1001}\:{cases},\:{i}\:{don}'{t}\:{think} \\ $$$${we}\:{can}\:{solve}\:{in}\:{this}\:{way}.\:{but}\:{i}\:{know} \\ $$$${no}\:{other}\:{ways}\:{how}\:{to}\:{solve}. \\ $$

Answered by mindispower last updated on 07/Dec/20

sonce rher is at least 2 in row we can use power serie (x^2 /(1−x)) starte withe x^2 ((x^2 /(1−x)))^5 , than find coeficient of x^(20) (x^(10) /((1−x)^5 )),(1/(1−x))=Σx^k ((1/(1−x)))^5 =(1/(4!)).(∂^4 /∂x^4 )((1/(1−x)))=(1/(4!)).Σ_(k≥0) (k+1)(k+2)(k+3)(k+4)x^k (x^(10) /((1−x)^5 ))=(1/(4!))Σ_(k≥0) (k+1)(k+2)(k+3)(k+4)x^(k+10) x^(20) hase coeficient (1/(4!)).(14)(13)(12)(11) =(((14)!)/(4!.10!))=C_(10) ^4

$${sonce}\:{rher}\:{is}\:{at}\:{least}\:\mathrm{2}\:{in}\:{row}\: \\ $$$${we}\:{can}\:{use}\:\:{power}\:{serie}\: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}}\:{starte}\:{withe}\:{x}^{\mathrm{2}} \\ $$$$\left(\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}}\right)^{\mathrm{5}} ,\:{than}\:{find}\:\boldsymbol{{coeficient}}\:\boldsymbol{{of}}\:{x}^{\mathrm{20}} \: \\ $$$$\frac{\boldsymbol{{x}}^{\mathrm{10}} }{\left(\mathrm{1}−\boldsymbol{{x}}\right)^{\mathrm{5}} },\frac{\mathrm{1}}{\mathrm{1}−{x}}=\Sigma{x}^{{k}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)^{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{4}!}.\frac{\partial^{\mathrm{4}} }{\partial{x}^{\mathrm{4}} }\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\frac{\mathrm{1}}{\mathrm{4}!}.\underset{{k}\geqslant\mathrm{0}} {\sum}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)\left({k}+\mathrm{4}\right){x}^{{k}} \\ $$$$\frac{{x}^{\mathrm{10}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{4}!}\underset{{k}\geqslant\mathrm{0}} {\sum}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)\left({k}+\mathrm{4}\right){x}^{{k}+\mathrm{10}} \\ $$$${x}^{\mathrm{20}} {hase}\:{coeficient}\:\frac{\mathrm{1}}{\mathrm{4}!}.\left(\mathrm{14}\right)\left(\mathrm{13}\right)\left(\mathrm{12}\right)\left(\mathrm{11}\right) \\ $$$$=\frac{\left(\mathrm{14}\right)!}{\mathrm{4}!.\mathrm{10}!}={C}_{\mathrm{10}} ^{\mathrm{4}} \\ $$

Commented by mr W last updated on 07/Dec/20

you had a typo. you meant C_(14) ^4 =1001. the difficulty is for example, if in the first row there are 5 students. to select 5 from 20 students there are C_5 ^(20) ways and to arrange these 5 students in the row there are 5! ways.

$${you}\:{had}\:{a}\:{typo}.\:{you}\:{meant}\:{C}_{\mathrm{14}} ^{\mathrm{4}} =\mathrm{1001}. \\ $$$${the}\:{difficulty}\:{is}\:{for}\:{example},\:{if}\:{in}\:{the} \\ $$$${first}\:{row}\:{there}\:{are}\:\mathrm{5}\:{students}.\:{to} \\ $$$${select}\:\mathrm{5}\:{from}\:\mathrm{20}\:{students}\:{there}\:{are} \\ $$$${C}_{\mathrm{5}} ^{\mathrm{20}} \:{ways}\:{and}\:{to}\:{arrange}\:{these}\:\mathrm{5} \\ $$$${students}\:{in}\:{the}\:{row}\:{there}\:{are}\:\mathrm{5}!\:{ways}. \\ $$

Answered by mr W last updated on 07/Dec/20

say the numbers of students in the five rows are a,b,c,d and e. a+b+c+d+e=20 2≤a,b,c,d,e≤12 to select a students for row 1 there are C_a ^(20) ways, and to arrange these a students in row 1 there are a! ways. the generating function for row 1 is Σ_(a≥2) C_a ^(20) a!x^a =Σ_(a=2) ^(12) P_a ^(20) x^a etc. GF=(Σ_(a=2) ^(12) P_a ^(20) x^a )(Σ_(b=2) ^(12) P_b ^(20−a) x^b )(Σ_(c=2) ^(12) P_c ^(20−a−b) x^c )(Σ_(d=2) ^(12) P_d ^(20−a−b−c) x^d )(Σ_(e=2) ^(12) P_e ^e x^e ) now it is to find the coefficient of the x^(20) term in GF which is Σ_(a+b+c+d+e=20_(a,b,c,d,e≥2) ) (P_a ^(20) P_b ^(20−a) P_c ^(20−a−b) P_d ^(20−a−b−c) P_e ^e )

$${say}\:{the}\:{numbers}\:{of}\:{students}\:{in}\:{the} \\ $$$${five}\:{rows}\:{are}\:{a},{b},{c},{d}\:{and}\:{e}. \\ $$$${a}+{b}+{c}+{d}+{e}=\mathrm{20} \\ $$$$\mathrm{2}\leqslant{a},{b},{c},{d},{e}\leqslant\mathrm{12} \\ $$$${to}\:{select}\:{a}\:{students}\:{for}\:{row}\:\mathrm{1}\:{there} \\ $$$${are}\:{C}_{{a}} ^{\mathrm{20}} \:{ways},\:{and}\:{to}\:{arrange}\:{these} \\ $$$${a}\:{students}\:{in}\:{row}\:\mathrm{1}\:{there}\:{are}\:{a}!\:{ways}. \\ $$$${the}\:{generating}\:{function}\:{for}\:{row}\:\mathrm{1}\:{is} \\ $$$$\underset{{a}\geqslant\mathrm{2}} {\sum}{C}_{{a}} ^{\mathrm{20}} {a}!{x}^{{a}} =\underset{{a}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{a}} ^{\mathrm{20}} {x}^{{a}} \\ $$$${etc}. \\ $$$${GF}=\left(\underset{{a}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{a}} ^{\mathrm{20}} {x}^{{a}} \right)\left(\underset{{b}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{b}} ^{\mathrm{20}−{a}} {x}^{{b}} \right)\left(\underset{{c}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{c}} ^{\mathrm{20}−{a}−{b}} {x}^{{c}} \right)\left(\underset{{d}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{d}} ^{\mathrm{20}−{a}−{b}−{c}} {x}^{{d}} \right)\left(\underset{{e}=\mathrm{2}} {\overset{\mathrm{12}} {\sum}}{P}_{{e}} ^{{e}} {x}^{{e}} \right) \\ $$$${now}\:{it}\:{is}\:{to}\:{find}\:{the}\:{coefficient}\:{of} \\ $$$${the}\:{x}^{\mathrm{20}} \:{term}\:{in}\:{GF}\:{which}\:{is} \\ $$$$\underset{\underset{{a},{b},{c},{d},{e}\geqslant\mathrm{2}} {{a}+{b}+{c}+{d}+{e}=\mathrm{20}}} {\sum}\left(\mathrm{P}_{{a}} ^{\mathrm{20}} \mathrm{P}_{{b}} ^{\mathrm{20}−{a}} \mathrm{P}_{{c}} ^{\mathrm{20}−{a}−{b}} \mathrm{P}_{{d}} ^{\mathrm{20}−{a}−{b}−{c}} \mathrm{P}_{{e}} ^{{e}} \right) \\ $$

Commented by mindispower last updated on 08/Dec/20

i see what you mean by your quation sorry im not in english is the place in row by somoene depend?

$${i}\:{see}\:{what}\:{you}\:{mean}\:{by}\:{your}\:{quation} \\ $$$${sorry}\:{im}\:{not}\:{in}\:{english}\:\:{is}\:{the}\:{place}\:{in}\:{row}\:{by}\:{somoene}\:{depend}? \\ $$

Commented by mr W last updated on 08/Dec/20

the five rows are distinct.

$${the}\:{five}\:{rows}\:{are}\:{distinct}. \\ $$

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