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20-students-should-stand-in-5-different-rows-each-row-should-have-at-least-2-students-in-how-many-ways-can-you-arrange-them-an-unsolved-old-question-

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Question Number 179948 by mr W last updated on 04/Nov/22
20 students should stand in 5 different rows. each row should have at least 2 students. in how many ways can you arrange them? (an unsolved old question)
$$\mathrm{20}\:{students}\:{should}\:{stand}\:{in}\:\mathrm{5} \\ $$$${different}\:{rows}.\:{each}\:{row}\:{should}\:{have} \\ $$$${at}\:{least}\:\mathrm{2}\:{students}.\:{in}\:{how}\:{many}\:{ways} \\ $$$${can}\:{you}\:{arrange}\:{them}? \\ $$$$\left({an}\:{unsolved}\:{old}\:{question}\right) \\ $$
Commented by mr W last updated on 04/Nov/22
for test purpose we can take 10 students in 3 rows.
$${for}\:{test}\:{purpose}\:{we}\:{can}\:{take}\:\mathrm{10}\: \\ $$$${students}\:{in}\:\mathrm{3}\:{rows}. \\ $$
Commented by JDamian last updated on 04/Nov/22
are the students distinguishable?
Commented by mr W last updated on 04/Nov/22
yes, we don′t have identical students.
$${yes},\:{we}\:{don}'{t}\:{have}\:{identical}\:{students}. \\ $$
Answered by Frix last updated on 04/Nov/22
5 rows 2 2 2 2 12 ×((5!)/(4!)) 2 2 2 3 11 ×((5!)/(3!)) 2 2 2 4 10 ×((5!)/(3!)) 2 2 2 5 9 ×((5!)/(3!)) 2 2 2 6 8 ×((5!)/(3!)) 2 2 2 7 7 ×((5!)/(3!2!)) 2 2 3 3 10 ×((5!)/(2!^2 )) 2 2 3 4 9 ×((5!)/(2!)) 2 2 3 5 8 ×((5!)/(2!)) 2 2 3 6 7 ×((5!)/(2!)) 2 2 4 4 8 ×((5!)/(2!^2 )) 2 2 4 5 7 ×((5!)/(2!)) 2 2 4 6 6 ×((5!)/(2!^2 )) 2 2 5 5 6 ×((5!)/(2!^2 )) 2 3 3 3 9 ×((5!)/(3!)) 2 3 3 4 8 ×((5!)/(2!)) 2 3 3 5 7 ×((5!)/(2!)) 2 3 3 6 6 ×((5!)/(2!^2 )) 2 3 4 4 7 ×((5!)/(2!)) 2 3 4 5 6 ×5! 2 3 5 5 5 ×((5!)/(3!)) 2 4 4 4 6 ×((5!)/(3!)) 2 4 4 5 5 ×((5!)/(2!^2 )) 3 3 3 3 8 ×((5!)/(4!)) 3 3 3 4 7 ×((5!)/(3!)) 3 3 3 5 6 ×((5!)/(3!)) 3 3 4 4 6 ×((5!)/(2!^2 )) 3 3 4 5 5 ×((5!)/(2!^2 )) 3 4 4 4 5 ×((5!)/(3!)) 4 4 4 4 4 ×((5!)/(5!)) 1001 possibilities 20 students = 20! possibilities total: 1001×20!=2 435 334 910 184 816 640 000
$$\mathrm{5}\:\mathrm{rows} \\ $$$$\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{12}\:×\frac{\mathrm{5}!}{\mathrm{4}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{11}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{4}\:\mathrm{10}\:×\frac{\mathrm{5}!}{\mathrm{3}!} \\ $$$$\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{5}\:\mathrm{9}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{6}\:\mathrm{8}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{7}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{3}!\mathrm{2}!} \\ $$$$\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{3}\:\mathrm{10}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{9}\:×\frac{\mathrm{5}!}{\mathrm{2}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{5}\:\mathrm{8}\:×\frac{\mathrm{5}!}{\mathrm{2}!} \\ $$$$\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{6}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{2}!}\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{4}\:\mathrm{4}\:\mathrm{8}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{4}\:\mathrm{5}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{2}!} \\ $$$$\mathrm{2}\:\mathrm{2}\:\mathrm{4}\:\mathrm{6}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{2}\:\mathrm{2}\:\mathrm{5}\:\mathrm{5}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{3}\:\mathrm{3}\:\mathrm{9}\:×\frac{\mathrm{5}!}{\mathrm{3}!} \\ $$$$\mathrm{2}\:\mathrm{3}\:\mathrm{3}\:\mathrm{4}\:\mathrm{8}\:×\frac{\mathrm{5}!}{\mathrm{2}!}\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{3}\:\mathrm{5}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{2}!}\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{3}\:\mathrm{6}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} } \\ $$$$\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{2}!}\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:×\mathrm{5}!\:\:\:\:\:\mathrm{2}\:\mathrm{3}\:\mathrm{5}\:\mathrm{5}\:\mathrm{5}\:×\frac{\mathrm{5}!}{\mathrm{3}!} \\ $$$$\mathrm{2}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{2}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\mathrm{5}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{3}\:\mathrm{3}\:\mathrm{3}\:\mathrm{3}\:\mathrm{8}\:×\frac{\mathrm{5}!}{\mathrm{4}!} \\ $$$$\mathrm{3}\:\mathrm{3}\:\mathrm{3}\:\mathrm{4}\:\mathrm{7}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{3}\:\mathrm{3}\:\mathrm{3}\:\mathrm{5}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{3}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{6}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} } \\ $$$$\mathrm{3}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{5}\:×\frac{\mathrm{5}!}{\mathrm{2}!^{\mathrm{2}} }\:\:\:\:\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:×\frac{\mathrm{5}!}{\mathrm{3}!}\:\:\:\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:×\frac{\mathrm{5}!}{\mathrm{5}!} \\ $$$$\mathrm{1001}\:\mathrm{possibilities} \\ $$$$\mathrm{20}\:\mathrm{students}\:=\:\mathrm{20}!\:\mathrm{possibilities} \\ $$$$\mathrm{total}: \\ $$$$\mathrm{1001}×\mathrm{20}!=\mathrm{2}\:\mathrm{435}\:\mathrm{334}\:\mathrm{910}\:\mathrm{184}\:\mathrm{816}\:\mathrm{640}\:\mathrm{000} \\ $$
Commented by mr W last updated on 05/Nov/22
yes, it′s correct, thanks sir!
$${yes},\:{it}'{s}\:{correct},\:{thanks}\:{sir}! \\ $$
Commented by mr W last updated on 05/Nov/22
the number is so terribly huge! the question is the same as in how many ways we can arrange 20 books in a book rack with 5 shelves. say we can work very quickly and need only one second for each arrangement, and we began to work as the universe began to exist, i.e. as the big bang occurred. what do you think? have we done all the possible arrangements? no! no by far! in fact we haven′t even done 0.02% of the work yet! not to think, what when we have 100 books!
$${the}\:{number}\:{is}\:{so}\:{terribly}\:{huge}! \\ $$$${the}\:{question}\:{is}\:{the}\:{same}\:{as}\:{in}\:{how} \\ $$$${many}\:{ways}\:{we}\:{can}\:{arrange}\:\mathrm{20}\:{books} \\ $$$${in}\:{a}\:{book}\:{rack}\:{with}\:\mathrm{5}\:{shelves}. \\ $$$${say}\:{we}\:{can}\:{work}\:{very}\:{quickly}\:{and} \\ $$$${need}\:{only}\:{one}\:{second}\:{for}\:{each}\: \\ $$$${arrangement},\:{and}\:{we}\:{began}\:{to}\:{work} \\ $$$${as}\:{the}\:{universe}\:{began}\:{to}\:{exist},\:{i}.{e}. \\ $$$${as}\:{the}\:{big}\:{bang}\:{occurred}.\:{what}\:{do}\:{you} \\ $$$${think}?\:{have}\:{we}\:{done}\:{all}\:{the}\:{possible} \\ $$$${arrangements}?\: \\ $$$${no}!\:{no}\:{by}\:{far}!\:{in}\:{fact}\:{we}\:{haven}'{t}\:{even} \\ $$$${done}\:\mathrm{0}.\mathrm{02\%}\:{of}\:{the}\:{work}\:{yet}! \\ $$$${not}\:{to}\:{think},\:{what}\:{when}\:{we}\:{have}\: \\ $$$$\mathrm{100}\:{books}! \\ $$
Commented by mr W last updated on 05/Nov/22
Answered by mr W last updated on 05/Nov/22
say the rows are labeled as A,B,C,D,E. the numbers of students in them are a,b,c,d,e respectively. a+b+c+d+e=20 2≤a,b,c,d,e the number of ways to distribute 20 people into 5 groups with a,b,c,d,e people respectively is: ((20!)/(a!b!c!d!e!)) to arrange the people in their rows there are a!b!c!d!e! ways. ((20!)/(a!b!c!d!e!))×a!b!c!d!e!=20! so we get the generating function 20!(x^2 +x^3 +...)^5 . the answer of the question is the coef. of x^(20) in the expansion of 20!(x^2 +x^3 +x^4 +...)^5 =((20!x^(10) )/((1−x)^5 )) =20!x^(10) Σ_(k=0) ^∞ C_4 ^(k+4) x^k 10+k=20 ⇒k=10 so the coef. of x^(20) is 20!C_4 ^(14) . generally for n people in r rows the answer is n!C_(r−1) ^(n−r−1) . in case of 10 people in 3 rows, the answer is 10!C_2 ^6 =54 432 000 this can be manually checked: 6+2+2 ⇒((10!)/(6!(2!)^2 2!))×3!×6!×2!×2!=((10!3!)/(2!)) 5+3+2 ⇒((10!)/(5!3!2!))×3!×5!×3!×2!=10!×3! 4+4+2 ⇒((10!)/((4!)^2 2!2!))×3!×4!×4!×2!=((10!3!)/(2!)) 4+3+3 ⇒((10!)/(4!(3!)^2 2!))×3!×4!×3!×3!=((10!3!)/(2!)) ⇒3×((10!3!)/(2!))+10!3!=54 432 000 ✓
$${say}\:{the}\:{rows}\:{are}\:{labeled}\:{as}\:{A},{B},{C},{D},{E}. \\ $$$${the}\:{numbers}\:{of}\:{students}\:{in}\:{them} \\ $$$${are}\:{a},{b},{c},{d},{e}\:{respectively}. \\ $$$${a}+{b}+{c}+{d}+{e}=\mathrm{20} \\ $$$$\mathrm{2}\leqslant{a},{b},{c},{d},{e} \\ $$$${the}\:{number}\:{of}\:{ways}\:{to}\:{distribute}\:\mathrm{20}\: \\ $$$${people}\:{into}\:\mathrm{5}\:{groups}\:{with}\:{a},{b},{c},{d},{e}\: \\ $$$${people}\:{respectively}\:{is}: \\ $$$$\frac{\mathrm{20}!}{{a}!{b}!{c}!{d}!{e}!} \\ $$$${to}\:{arrange}\:{the}\:{people}\:{in}\:{their}\:{rows} \\ $$$${there}\:{are}\:{a}!{b}!{c}!{d}!{e}!\:{ways}. \\ $$$$\frac{\mathrm{20}!}{{a}!{b}!{c}!{d}!{e}!}×{a}!{b}!{c}!{d}!{e}!=\mathrm{20}! \\ $$$${so}\:{we}\:{get}\:{the}\:{generating}\:{function} \\ $$$$\mathrm{20}!\left({x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{5}} .\:{the}\:{answer}\:{of}\:{the} \\ $$$${question}\:{is}\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{20}} \:{in}\:{the} \\ $$$${expansion}\:{of}\: \\ $$$$\mathrm{20}!\left({x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +…\right)^{\mathrm{5}} \\ $$$$=\frac{\mathrm{20}!{x}^{\mathrm{10}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} } \\ $$$$=\mathrm{20}!{x}^{\mathrm{10}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}} \\ $$$$\mathrm{10}+{k}=\mathrm{20}\:\Rightarrow{k}=\mathrm{10} \\ $$$${so}\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{20}} \:{is}\:\mathrm{20}!{C}_{\mathrm{4}} ^{\mathrm{14}} . \\ $$$${generally}\:{for}\:{n}\:{people}\:{in}\:{r}\:{rows}\:{the} \\ $$$${answer}\:{is}\:{n}!{C}_{{r}−\mathrm{1}} ^{{n}−{r}−\mathrm{1}} . \\ $$$$ \\ $$$${in}\:{case}\:{of}\:\mathrm{10}\:{people}\:{in}\:\mathrm{3}\:{rows},\:{the} \\ $$$${answer}\:{is} \\ $$$$\mathrm{10}!{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{54}\:\mathrm{432}\:\mathrm{000} \\ $$$${this}\:{can}\:{be}\:{manually}\:{checked}: \\ $$$$\mathrm{6}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{10}!}{\mathrm{6}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{3}!×\mathrm{6}!×\mathrm{2}!×\mathrm{2}!=\frac{\mathrm{10}!\mathrm{3}!}{\mathrm{2}!} \\ $$$$\mathrm{5}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{10}!}{\mathrm{5}!\mathrm{3}!\mathrm{2}!}×\mathrm{3}!×\mathrm{5}!×\mathrm{3}!×\mathrm{2}!=\mathrm{10}!×\mathrm{3}! \\ $$$$\mathrm{4}+\mathrm{4}+\mathrm{2}\:\Rightarrow\frac{\mathrm{10}!}{\left(\mathrm{4}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{3}!×\mathrm{4}!×\mathrm{4}!×\mathrm{2}!=\frac{\mathrm{10}!\mathrm{3}!}{\mathrm{2}!} \\ $$$$\mathrm{4}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{10}!}{\mathrm{4}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{3}!×\mathrm{4}!×\mathrm{3}!×\mathrm{3}!=\frac{\mathrm{10}!\mathrm{3}!}{\mathrm{2}!} \\ $$$$\Rightarrow\mathrm{3}×\frac{\mathrm{10}!\mathrm{3}!}{\mathrm{2}!}+\mathrm{10}!\mathrm{3}!=\mathrm{54}\:\mathrm{432}\:\mathrm{000}\:\checkmark \\ $$
Commented by mr W last updated on 05/Nov/22
alternative way instead of using generating function: a+b+c+d+e=20 2≤a,b,c,d,e let a=a′+1, b=b′+1, ... a′+b′+c′+d′+e′=15 1≤a′,b′,c′,d′,e′ ★∣★∣★∣★∣★∣★∣★...★∣★∣★ (15 stars and 14 bars) the number of ways to select 4 of the 14 bars is C_4 ^(14) =1001. so the answer is 20!C_4 ^(14) =1001×20!
$${alternative}\:{way}\:{instead}\:{of}\:{using} \\ $$$${generating}\:{function}: \\ $$$${a}+{b}+{c}+{d}+{e}=\mathrm{20} \\ $$$$\mathrm{2}\leqslant{a},{b},{c},{d},{e} \\ $$$${let}\:{a}={a}'+\mathrm{1},\:{b}={b}'+\mathrm{1},\:… \\ $$$${a}'+{b}'+{c}'+{d}'+{e}'=\mathrm{15} \\ $$$$\mathrm{1}\leqslant{a}',{b}',{c}',{d}',{e}' \\ $$$$\bigstar\mid\bigstar\mid\bigstar\mid\bigstar\mid\bigstar\mid\bigstar\mid\bigstar…\bigstar\mid\bigstar\mid\bigstar \\ $$$$\left(\mathrm{15}\:{stars}\:{and}\:\mathrm{14}\:{bars}\right) \\ $$$${the}\:{number}\:{of}\:{ways}\:{to}\:{select}\:\mathrm{4}\:{of}\:{the} \\ $$$$\mathrm{14}\:{bars}\:{is}\:{C}_{\mathrm{4}} ^{\mathrm{14}} =\mathrm{1001}.\:{so}\:{the}\:{answer}\:{is} \\ $$$$\mathrm{20}!{C}_{\mathrm{4}} ^{\mathrm{14}} =\mathrm{1001}×\mathrm{20}! \\ $$
Commented by Frix last updated on 05/Nov/22
thank you for explaining
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{explaining} \\ $$

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