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Question Number 179948 by mr W last updated on 04/Nov/22

$$20studentsshouldstandin5$$$$differentrows.eachrowshouldhave$$$$atleast2students.inhowmanyways$$$$canyouarrangethem?$$$$\left(anunsolvedoldquestion\right)$$

Commented by mr W last updated on 04/Nov/22

$$fortestpurposewecantake10$$$$studentsin3rows.$$

Commented by JDamian last updated on 04/Nov/22

are the students distinguishable?

Commented by mr W last updated on 04/Nov/22

$$yes,we{don}^{\prime }thaveidenticalstudents.$$

Answered by Frix last updated on 04/Nov/22

$$5\mathrm{rows}$$$$222212\times \frac{5!}{4!}222311\times \frac{5!}{3!}222410\times \frac{5!}{3!}$$$$22259\times \frac{5!}{3!}22268\times \frac{5!}{3!}22277\times \frac{5!}{3!2!}$$$$223310\times \frac{5!}{2{!}^2}22349\times \frac{5!}{2!}22358\times \frac{5!}{2!}$$$$22367\times \frac{5!}{2!}22448\times \frac{5!}{2{!}^2}22457\times \frac{5!}{2!}$$$$22466\times \frac{5!}{2{!}^2}22556\times \frac{5!}{2{!}^2}23339\times \frac{5!}{3!}$$$$23348\times \frac{5!}{2!}23357\times \frac{5!}{2!}23366\times \frac{5!}{2{!}^2}$$$$23447\times \frac{5!}{2!}23456\times 5!23555\times \frac{5!}{3!}$$$$24446\times \frac{5!}{3!}24455\times \frac{5!}{2{!}^2}33338\times \frac{5!}{4!}$$$$33347\times \frac{5!}{3!}33356\times \frac{5!}{3!}33446\times \frac{5!}{2{!}^2}$$$$33455\times \frac{5!}{2{!}^2}34445\times \frac{5!}{3!}44444\times \frac{5!}{5!}$$$$1001\mathrm{possibilities}$$$$20\mathrm{students}=20!\mathrm{possibilities}$$$$\mathrm{total}:$$$$1001\times 20!=2435334910184816640000$$

Commented by mr W last updated on 05/Nov/22

$$yes,{it}^{\prime }scorrect,thankssir!$$

Commented by mr W last updated on 05/Nov/22

$$thenumberissoterriblyhuge!$$$$thequestionisthesameasinhow$$$$manywayswecanarrange20books$$$$inabookrackwith5shelves.$$$$saywecanworkveryquicklyand$$$$needonlyonesecondforeach$$$$arrangement,andwebegantowork$$$$astheuniversebegantoexist,i.e.$$$$asthebigbangoccurred.whatdoyou$$$$think?havewedoneallthepossible$$$$arrangements?$$$$no!nobyfar!infactwe{haven}^{\prime }teven$$$$done0.02\mathrm{\%}oftheworkyet!$$$$nottothink,whatwhenwehave$$$$100books!$$

Commented by mr W last updated on 05/Nov/22

Answered by mr W last updated on 05/Nov/22

$$saytherowsarelabeledasA,B,C,D,E.$$$$thenumbersofstudentsinthem$$$$area,b,c,d,erespectively.$$$$a+b+c+d+e=20$$$$2⩽a,b,c,d,e$$$$thenumberofwaystodistribute20$$$$peopleinto5groupswitha,b,c,d,e$$$$peoplerespectivelyis:$$$$\frac{20!}{a!b!c!d!e!}$$$$toarrangethepeopleintheirrows$$$$therearea!b!c!d!e!ways.$$$$\frac{20!}{a!b!c!d!e!}\times a!b!c!d!e!=20!$$$$sowegetthegeneratingfunction$$$$20!{(x^2+x^3+...)}^5.theanswerofthe$$$$questionisthecoef.of{x}^{20}inthe$$$$expansionof$$$$20!{(x^2+x^3+x^4+...)}^5$$$$=\frac{20!x^{10}}{{(1-x)}^5}$$$$=20!x^{10}\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_4^{k+4}{x}^k$$$$10+k=20\Rightarrow k=10$$$$sothecoef.of{x}^{20}is20!C_4^{14}.$$$$generallyfornpeopleinrrowsthe$$$$answerisn!C_{r-1}^{n-r-1}.$$$$$$$$incaseof10peoplein3rows,the$$$$answeris$$$$10!C_2^6=54432000$$$$thiscanbemanuallychecked:$$$$6+2+2\Rightarrow \frac{10!}{6!{(2!)}^{2}2!}\times 3!\times 6!\times 2!\times 2!=\frac{10!3!}{2!}$$$$5+3+2\Rightarrow \frac{10!}{5!3!2!}\times 3!\times 5!\times 3!\times 2!=10!\times 3!$$$$4+4+2\Rightarrow \frac{10!}{{(4!)}^{2}2!2!}\times 3!\times 4!\times 4!\times 2!=\frac{10!3!}{2!}$$$$4+3+3\Rightarrow \frac{10!}{4!{(3!)}^{2}2!}\times 3!\times 4!\times 3!\times 3!=\frac{10!3!}{2!}$$$$\Rightarrow 3\times \frac{10!3!}{2!}+10!3!=54432000✓$$

Commented by mr W last updated on 05/Nov/22

$$alternativewayinsteadofusing$$$$generatingfunction:$$$$a+b+c+d+e=20$$$$2⩽a,b,c,d,e$$$$leta=a^{\prime }+1,b=b^{\prime }+1,...$$$$a^{\prime }+b^{\prime }+c^{\prime }+d^{\prime }+e^{\prime }=15$$$$1⩽a^{\prime },b^{\prime },c^{\prime },d^{\prime },e^{\prime }$$$$★\mid ★\mid ★\mid ★\mid ★\mid ★\mid ★...★\mid ★\mid ★$$$$\left(15starsand14bars\right)$$$$thenumberofwaystoselect4ofthe$$$$14barsis{C}_4^{14}=1001.sotheansweris$$$$20!C_4^{14}=1001\times 20!$$

Commented by Frix last updated on 05/Nov/22

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{explaining}$$

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