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Question Number 179818 by mr W last updated on 02/Nov/22

in how many ways can you distribute 20 different books to 5 students such that each student gets at least 2 books?

$${in}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{distribute} \\ $$$$\mathrm{20}\:{different}\:{books}\:{to}\:\mathrm{5}\:{students}\:{such} \\ $$$${that}\:{each}\:{student}\:{gets}\:{at}\:{least}\:\mathrm{2}\:{books}? \\ $$

Answered by a.lgnaoui last updated on 03/Nov/22

3 4 4 4 5 4 3 4 4 5 3 4 4 5 4 4 3 4 5 4 3 4 5 4 4 4 3 5 4 4 3 5 4 4 4 5 3 4 4 4 4 4 3 4 5 4 4 4 3 5 4 4 3 5 4 4 4 5 3 4 4 5 3 4 4 4 5 4 3 4 5 4 3 4 4 5 4 4 3 4 4 4 4 5 3 4 4 5 4 3 total=20ways 4 5 4 4 3 5 4 4 4 3

$$ \\ $$$$\:\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\:\:\:\mathrm{4}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5} \\ $$$$\:\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\mathrm{4}\:\:\:\:\mathrm{4}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{4} \\ $$$$\:\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{4}\:\mathrm{4}\:\:\:\:\:\mathrm{4}\:\mathrm{3}\:\mathrm{5}\:\mathrm{4}\:\mathrm{4} \\ $$$$\:\:\mathrm{3}\:\mathrm{5}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\:\:\:\:\mathrm{5}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4} \\ $$$$ \\ $$$$\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\:\:\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{3}\:\mathrm{5} \\ $$$$\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{3}\:\mathrm{5}\:\mathrm{4}\:\:\:\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\mathrm{3}\:\mathrm{4} \\ $$$$\:\:\mathrm{4}\:\mathrm{5}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\:\:\:\:\mathrm{4}\:\mathrm{5}\:\mathrm{4}\:\mathrm{3}\:\mathrm{4} \\ $$$$\:\:\mathrm{5}\:\mathrm{4}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}\:\:\:\:\:\mathrm{5}\:\mathrm{4}\:\mathrm{4}\:\mathrm{3}\:\mathrm{4} \\ $$$$ \\ $$$$\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\mathrm{3} \\ $$$$\:\:\mathrm{4}\:\mathrm{4}\:\mathrm{5}\:\mathrm{4}\:\mathrm{3}\:\:\:\:\:{total}=\mathrm{20}{ways} \\ $$$$\:\:\mathrm{4}\:\mathrm{5}\:\mathrm{4}\:\mathrm{4}\:\mathrm{3} \\ $$$$\:\:\mathrm{5}\:\mathrm{4}\:\mathrm{4}\:\mathrm{4}\:\mathrm{3} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 03/Nov/22

in my answer I considered no difference betwen books order 1 2 3 4 5 (for sudents) 3,4,5 Number of books for each student.

$$\:{in}\:{my}\:{answer}\:{I}\:{considered} \\ $$$${no}\:{difference}\:{betwen}\:{books} \\ $$$${order}\:\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\:\:\left({for}\:{sudents}\right) \\ $$$$\mathrm{3},\mathrm{4},\mathrm{5}\:\:\:{Number}\:{of}\:{books}\:{for} \\ $$$${each}\:{student}. \\ $$

Commented by mr W last updated on 03/Nov/22

at least 2 books ≠ more than 2 books besides the books are different and the students are also different (logically!)

$${at}\:{least}\:\mathrm{2}\:{books}\:\neq\:{more}\:{than}\:\mathrm{2}\:{books} \\ $$$${besides}\:{the}\:{books}\:{are}\:{different}\:{and} \\ $$$${the}\:{students}\:{are}\:{also}\:{different}\: \\ $$$$\left({logically}!\right) \\ $$

Answered by Acem last updated on 04/Nov/22

Primitive Solution draft A) distributing 2 dif.bks. to each stu. A= C_2 ^( 20) C_2 ^( 18) C_2 ^( 16) C_2 ^( 14) C_2 ^( 12) 5!= 2 514 159 648 000 B) distributing the other 10 dif.bks. to 5 stu. B= C_(10) ^( 10) ×5 + C_9 ^( 10) ×5. C_1 ^( 1) ×4 + C_8 ^( 10) ×5 C_2 ^( 2) ×4 .... + C_2 ^( 10) ×5 C_8 ^( 8) ×4 + C_1 ^( 10) ×5 C_9 ^( 9) ×4 NumWays= A+B

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Primitive}}\:\boldsymbol{{Solution}}\:\boldsymbol{{draft}} \\ $$$$ \\ $$$$\left.{A}\right)\:{distributing}\:\mathrm{2}\:{dif}.{bks}.\:{to}\:{each}\:{stu}. \\ $$$$\:{A}=\:{C}_{\mathrm{2}} ^{\:\mathrm{20}} \:{C}_{\mathrm{2}} ^{\:\mathrm{18}} \:{C}_{\mathrm{2}} ^{\:\mathrm{16}} \:{C}_{\mathrm{2}} ^{\:\mathrm{14}} \:{C}_{\mathrm{2}} ^{\:\mathrm{12}} \:\mathrm{5}!=\:\mathrm{2}\:\mathrm{514}\:\mathrm{159}\:\mathrm{648}\:\mathrm{000} \\ $$$$ \\ $$$$\left.{B}\right)\:{distributing}\:{the}\:{other}\:\mathrm{10}\:{dif}.{bks}.\:{to}\:\mathrm{5}\:{stu}. \\ $$$$\:{B}=\:{C}_{\mathrm{10}} ^{\:\mathrm{10}} \:×\mathrm{5}\:+\:{C}_{\mathrm{9}} ^{\:\mathrm{10}} ×\mathrm{5}.\:{C}_{\mathrm{1}} ^{\:\mathrm{1}} ×\mathrm{4}\:+\:{C}_{\mathrm{8}} ^{\:\mathrm{10}} ×\mathrm{5}\:{C}_{\mathrm{2}} ^{\:\mathrm{2}} ×\mathrm{4} \\ $$$$\:\:\:\:\:\:....\:+\:\:{C}_{\mathrm{2}} ^{\:\mathrm{10}} ×\mathrm{5}\:{C}_{\mathrm{8}} ^{\:\mathrm{8}} ×\mathrm{4}\:+\:\:{C}_{\mathrm{1}} ^{\:\mathrm{10}} ×\mathrm{5}\:{C}_{\mathrm{9}} ^{\:\mathrm{9}} ×\mathrm{4} \\ $$$$\:{NumWays}=\:{A}+{B} \\ $$

Answered by mr W last updated on 03/Nov/22

the number of ways to distribute 20 books to 5 students such that each student gets at least 2 books is the coef. of x^(20) in the expansion of 20!(e^x −1−x)^5 which is 64 341 775 001 400.

$${the}\:{number}\:{of}\:{ways}\:{to}\:{distribute}\:\mathrm{20} \\ $$$${books}\:{to}\:\mathrm{5}\:{students}\:{such}\:{that}\:{each} \\ $$$${student}\:{gets}\:{at}\:{least}\:\mathrm{2}\:{books}\:{is}\:{the} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{20}} \:{in}\:{the}\:{expansion}\:{of}\: \\ $$$$\mathrm{20}!\left({e}^{{x}} −\mathrm{1}−{x}\right)^{\mathrm{5}} \:{which}\:{is} \\ $$$$\mathrm{64}\:\mathrm{341}\:\mathrm{775}\:\mathrm{001}\:\mathrm{400}. \\ $$

Commented by mr W last updated on 03/Nov/22

Commented by mr W last updated on 03/Nov/22

i solved in a way which seems to be a good way to me. i can′t and won′t answer why i did it differently as you. my thought was: in step 1 i let each student get one book and in step 2 each gets at least one book. i considered again and found out that i can also solve the question with my method (using generating function) even without step 1. so i have modified my solution correspondingly. i couldn′t totally follow your solution, but i think it is not correct. you can check your solution for example that we distribute 5 books to 2 students and each gets at least 2 books. the correct answer should be 20. but your method gets already in step 1 A= C_2 ^( 5) C_2 ^( 3) 2!=60. that means the method is wrong. so please recheck your own solution.

$${i}\:{solved}\:{in}\:{a}\:{way}\:{which}\:{seems}\:{to}\:{be}\:{a} \\ $$$${good}\:{way}\:{to}\:{me}.\:{i}\:{can}'{t}\:{and}\:{won}'{t} \\ $$$${answer}\:{why}\:{i}\:{did}\:{it}\:{differently}\:{as}\: \\ $$$${you}. \\ $$$${my}\:{thought}\:{was}:\:{in}\:{step}\:\mathrm{1}\:{i}\:{let}\:{each} \\ $$$${student}\:{get}\:{one}\:{book}\:{and}\:{in}\:{step}\:\mathrm{2} \\ $$$${each}\:{gets}\:{at}\:{least}\:{one}\:{book}.\: \\ $$$${i}\:{considered}\:{again}\:{and}\:{found}\:{out}\:{that} \\ $$$${i}\:{can}\:{also}\:{solve}\:{the}\:{question}\:{with}\:{my}\: \\ $$$${method}\:\left({using}\:{generating}\:{function}\right)\: \\ $$$${even}\:{without}\:{step}\:\mathrm{1}.\:{so}\:{i}\:{have}\:{modified}\: \\ $$$${my}\:{solution}\:{correspondingly}. \\ $$$$ \\ $$$${i}\:{couldn}'{t}\:{totally}\:{follow}\:{your}\:{solution}, \\ $$$${but}\:{i}\:{think}\:{it}\:{is}\:{not}\:{correct}.\:{you}\:{can} \\ $$$${check}\:{your}\:{solution}\:{for}\:{example}\:{that} \\ $$$${we}\:{distribute}\:\mathrm{5}\:{books}\:{to}\:\mathrm{2}\:{students} \\ $$$${and}\:{each}\:{gets}\:{at}\:{least}\:\mathrm{2}\:{books}.\:{the} \\ $$$${correct}\:{answer}\:{should}\:{be}\:\mathrm{20}.\:{but}\:{your}\: \\ $$$${method}\:{gets}\:{already}\:{in}\:{step}\:\mathrm{1}\: \\ $$$${A}=\:{C}_{\mathrm{2}} ^{\:\mathrm{5}} \:{C}_{\mathrm{2}} ^{\:\mathrm{3}} \:\mathrm{2}!=\mathrm{60}.\:{that}\:{means}\:{the} \\ $$$${method}\:{is}\:{wrong}.\:{so}\:{please}\:{recheck} \\ $$$${your}\:{own}\:{solution}. \\ $$

Commented by Acem last updated on 03/Nov/22

Hello, For the 1st step, why only 5 books not 10 we can choose randomly 2books and give them to someone as follow A) distributing 2 dif.bks. to each stu. A= C_2 ^( 20) C_2 ^( 18) C_2 ^( 16) C_2 ^( 14) C_2 ^( 12) 5!= 2 514 159 648 000

$${Hello}, \\ $$$${For}\:{the}\:\mathrm{1}{st}\:{step},\:{why}\:{only}\:\mathrm{5}\:{books}\:{not}\:\mathrm{10} \\ $$$${we}\:{can}\:{choose}\:{randomly}\:\mathrm{2}{books}\:{and}\:{give} \\ $$$$\:{them}\:{to}\:{someone}\:{as}\:{follow} \\ $$$$\: \\ $$$$\left.{A}\right)\:{distributing}\:\mathrm{2}\:{dif}.{bks}.\:{to}\:{each}\:{stu}. \\ $$$$\:{A}=\:{C}_{\mathrm{2}} ^{\:\mathrm{20}} \:{C}_{\mathrm{2}} ^{\:\mathrm{18}} \:{C}_{\mathrm{2}} ^{\:\mathrm{16}} \:{C}_{\mathrm{2}} ^{\:\mathrm{14}} \:{C}_{\mathrm{2}} ^{\:\mathrm{12}} \:\mathrm{5}!=\:\mathrm{2}\:\mathrm{514}\:\mathrm{159}\:\mathrm{648}\:\mathrm{000} \\ $$$$ \\ $$

Commented by Acem last updated on 03/Nov/22

Thank you my friend for everything, in fact i am sure that i′ve something missing with the 2 step that is why didn′t compt it result, and will try to fix it later, but now am shocked about the 1st step, ok i will recheck it when finish my work. Have a nice time

$${Thank}\:{you}\:{my}\:{friend}\:{for}\:{everything},\:{in}\:{fact}\:{i}\:{am} \\ $$$$\:{sure}\:{that}\:{i}'{ve}\:{something}\:{missing}\:{with}\:{the}\:\mathrm{2}\:{step} \\ $$$$\:{that}\:{is}\:{why}\:{didn}'{t}\:{compt}\:{it}\:{result},\:{and}\:{will}\:{try} \\ $$$$\:{to}\:{fix}\:{it}\:{later},\:{but}\:{now}\:{am}\:{shocked}\:{about}\:{the}\:\mathrm{1}{st} \\ $$$$\:{step},\:{ok}\:{i}\:{will}\:{recheck}\:{it}\:{when}\:{finish}\:{my}\:{work}. \\ $$$$\:{Have}\:{a}\:{nice}\:{time} \\ $$

Commented by Acem last updated on 03/Nov/22

Hello again, I got the mistake A= C_2 ^( 20) C_2 ^( 18) C_2 ^( 16) C_2 ^( 14) “C_2 ^( 12) ” 5! and the true is: A= C_2 ^( 20) C_2 ^( 18) C_2 ^( 16) C_2 ^( 14) C_(12) ^( 12) 5!= 38 093 328 000 ✓ Now, dealing with 12, 11, 10, ... books is easier than dealing with 15, 14, ... books Anyway i got the result: 62 174 055 528 000 I will feed back my method very soon If yours is true that mean i still missing some cases of distrubtion P.s: please make the question with less number as 9 books for example as long as the idea of the solution is well known Thank you

$$ \\ $$$${Hello}\:{again}, \\ $$$$\:{I}\:{got}\:{the}\:{mistake} \\ $$$$\:{A}=\:{C}_{\mathrm{2}} ^{\:\mathrm{20}} \:{C}_{\mathrm{2}} ^{\:\mathrm{18}} \:{C}_{\mathrm{2}} ^{\:\mathrm{16}} \:{C}_{\mathrm{2}} ^{\:\mathrm{14}} \:``\boldsymbol{{C}}_{\mathrm{2}} ^{\:\mathrm{12}} ''\:\mathrm{5}! \\ $$$$\:{and}\:{the}\:{true}\:{is}: \\ $$$$\:{A}=\:{C}_{\mathrm{2}} ^{\:\mathrm{20}} \:{C}_{\mathrm{2}} ^{\:\mathrm{18}} \:{C}_{\mathrm{2}} ^{\:\mathrm{16}} \:{C}_{\mathrm{2}} ^{\:\mathrm{14}} \:\boldsymbol{{C}}_{\mathrm{12}} ^{\:\mathrm{12}} \:\mathrm{5}!=\:\mathrm{38}\:\mathrm{093}\:\mathrm{328}\:\mathrm{000}\:\checkmark \\ $$$${Now},\:{dealing}\:{with}\:\mathrm{12},\:\mathrm{11},\:\mathrm{10},\:...\:{books}\:{is}\:{easier} \\ $$$$\:{than}\:{dealing}\:{with}\:\mathrm{15},\:\mathrm{14},\:...\:{books} \\ $$$$ \\ $$$${Anyway}\:{i}\:{got}\:{the}\:{result}:\:\mathrm{62}\:\mathrm{174}\:\mathrm{055}\:\mathrm{528}\:\mathrm{000} \\ $$$${I}\:{will}\:{feed}\:{back}\:{my}\:{method}\:{very}\:{soon} \\ $$$$ \\ $$$$\:{If}\:{yours}\:{is}\:{true}\:{that}\:{mean}\:{i}\:{still}\:{missing}\:{some} \\ $$$$\:{cases}\:{of}\:{distrubtion} \\ $$$$ \\ $$$${P}.{s}:\:{please}\:{make}\:{the}\:{question}\:{with}\:{less}\:{number} \\ $$$$\:{as}\:\mathrm{9}\:{books}\:{for}\:{example}\:{as}\:{long}\:{as}\:{the}\:{idea}\:{of} \\ $$$$\:{the}\:{solution}\:{is}\:{well}\:{known} \\ $$$$ \\ $$$${Thank}\:{you} \\ $$$$ \\ $$

Commented by mr W last updated on 03/Nov/22

as i have said, you can check your method with the most simple case: 5 books to 2 student, at least 2 books for each student. i got the answer: 20 ways. this is true and can be easily proved.

$${as}\:{i}\:{have}\:{said},\:{you}\:{can}\:{check}\:{your} \\ $$$${method}\:{with}\:{the}\:{most}\:{simple}\:{case}: \\ $$$$\mathrm{5}\:{books}\:{to}\:\mathrm{2}\:{student},\:{at}\:{least}\:\mathrm{2}\:{books} \\ $$$${for}\:{each}\:{student}. \\ $$$${i}\:{got}\:{the}\:{answer}:\:\mathrm{20}\:{ways}.\:{this}\:{is} \\ $$$${true}\:{and}\:{can}\:{be}\:{easily}\:{proved}. \\ $$

Commented by Acem last updated on 03/Nov/22

I know that, but as the number of items increases the branches increase, and here if we neglect one case of distrubtion we would lose its branches

$${I}\:{know}\:{that},\:{but}\:{as}\:{the}\:{number}\:{of}\:{items}\:{increases} \\ $$$$\:{the}\:{branches}\:{increase},\:{and}\:{here}\:{if}\:{we}\:{neglect} \\ $$$$\:{one}\:{case}\:{of}\:{distrubtion}\:{we}\:{would}\:{lose}\:{its}\:{branches} \\ $$$$ \\ $$

Commented by mr W last updated on 03/Nov/22

then you can put an example by yourself which you think is good.

$${then}\:{you}\:{can}\:{put}\:{an}\:{example}\:{by} \\ $$$${yourself}\:{which}\:{you}\:{think}\:{is}\:{good}. \\ $$

Answered by Acem last updated on 04/Nov/22

The solve in one stage NumWays= 5! [C_(12) ^( 20) C_( 2) ^( 8) C_2 ^( 6) C_2 ^( 4) C_2 ^2 + C_(11) ^( 20) C_( 3) ^( 9) C_2 ^( 6) C_2 ^( 4) C_2 ^( 2) + C_(10) ^( 20) ( C_( 4) ^( 10) C_2 ^( 6) C_2 ^( 4) C_2 ^2 + C_( 3) ^( 10) C_( 3) ^( 7) C_2 ^( 4) C_2 ^( 2) + C_9 ^( 20) (C_( 5) ^( 11) C_2 ^( 6) C_2 ^( 4) C_2 ^2 + C_( 4) ^( 11) C_3 ^( 7) C_2 ^( 4) C_2 ^( 2) + + C_( 3) ^( 11) C_3 ^( 8) C_3 ^( 5) C_2 ^( 2) ) + C_( 8) ^( 20) (C_( 6) ^( 12) C_2 ^( 6) C_2 ^( 4) C_2 ^2 + C_( 5) ^( 12) C_( 3) ^( 7) C_2 ^( 4) C_2 ^( 2) + C_( 4) ^( 12) C_( 4) ^( 8) C_2 ^( 4) C_2 ^2 + C_( 4) ^( 12) C_( 3) ^( 8) C_( 3) ^( 5) C_2 ^( 2) + C_( 3) ^( 12) C_( 3) ^( 9) C_( 3) ^( 6) C_( 3) ^( 3) ) + C_( 7) ^( 20) (C_( 7) ^( 13) C_2 ^( 6) C_2 ^( 4) C_2 ^2 + C_( 6) ^( 13) C_( 3) ^( 7) C_2 ^( 4) C_2 ^( 2) + C_( 5) ^( 13) C_( 4) ^( 8) C_2 ^( 4) C_2 ^2 + C_( 5) ^( 13) C_( 3) ^( 8) C_( 3) ^( 5) C_2 ^( 2) C_( 4) ^( 13) C_( 4) ^( 9) C_( 3) ^( 5) C_2 ^( 2) + C_( 4) ^( 13) C_( 3) ^( 9) C_( 3) ^( 6) C_( 3) ^( 3) )] NumWays= 62 174 055 528 000 Things are very clear in front of you and so far i have not felt that i′ve overlooked one of the cases

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{The}}\:\boldsymbol{{solve}}\:\boldsymbol{{in}}\:\boldsymbol{{one}}\:\boldsymbol{{stage}} \\ $$$${NumWays}=\: \\ $$$$\:\:\:\:\:\mathrm{5}!\:\left[{C}_{\mathrm{12}} ^{\:\mathrm{20}} \:{C}_{\:\mathrm{2}} ^{\:\mathrm{8}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:+\:{C}_{\mathrm{11}} ^{\:\mathrm{20}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{9}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:+\right. \\ $$$$\:\:\:\:\:\:{C}_{\mathrm{10}} ^{\:\mathrm{20}} \:\left(\:{C}_{\:\mathrm{4}} ^{\:\mathrm{10}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:+\:{C}_{\:\mathrm{3}} ^{\:\mathrm{10}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{7}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:\:\:\:+\right. \\ $$$$\:\:\:\:\:\:{C}_{\mathrm{9}} ^{\:\mathrm{20}} \:\left({C}_{\:\mathrm{5}} ^{\:\mathrm{11}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:+\:{C}_{\:\mathrm{4}} ^{\:\mathrm{11}} \:{C}_{\mathrm{3}} ^{\:\mathrm{7}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:\:\:+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:{C}_{\:\mathrm{3}} ^{\:\mathrm{11}} \:{C}_{\mathrm{3}} ^{\:\mathrm{8}} \:{C}_{\mathrm{3}} ^{\:\mathrm{5}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\right)\:\:\:\:\:+ \\ $$$$\:\:\:\:{C}_{\:\mathrm{8}} ^{\:\mathrm{20}} \:\left({C}_{\:\mathrm{6}} ^{\:\mathrm{12}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:+\:{C}_{\:\mathrm{5}} ^{\:\mathrm{12}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{7}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:\:\:\:+\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{\:\mathrm{4}} ^{\:\mathrm{12}} \:{C}_{\:\mathrm{4}} ^{\:\mathrm{8}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:+\:{C}_{\:\mathrm{4}} ^{\:\mathrm{12}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{8}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{5}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:\:\:\:+ \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{\:\mathrm{3}} ^{\:\mathrm{12}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{9}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{3}} \:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+ \\ $$$$\:\:\:\:{C}_{\:\mathrm{7}} ^{\:\mathrm{20}} \:\left({C}_{\:\mathrm{7}} ^{\:\mathrm{13}} \:{C}_{\mathrm{2}} ^{\:\mathrm{6}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:+\:{C}_{\:\mathrm{6}} ^{\:\mathrm{13}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{7}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:\:\:\:+\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{\:\mathrm{5}} ^{\:\mathrm{13}} \:{C}_{\:\mathrm{4}} ^{\:\mathrm{8}} \:{C}_{\mathrm{2}} ^{\:\mathrm{4}} \:{C}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:+\:{C}_{\:\mathrm{5}} ^{\:\mathrm{13}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{8}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{5}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \: \\ $$$$\left.\:\left.\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{\:\mathrm{4}} ^{\:\mathrm{13}} \:{C}_{\:\mathrm{4}} ^{\:\mathrm{9}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{5}} \:{C}_{\mathrm{2}} ^{\:\mathrm{2}} \:\:\:\:\:+\:{C}_{\:\mathrm{4}} ^{\:\mathrm{13}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{9}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} \:{C}_{\:\mathrm{3}} ^{\:\:\mathrm{3}} \:\:\right)\right] \\ $$$$ \\ $$$${NumWays}=\:\mathrm{62}\:\mathrm{174}\:\mathrm{055}\:\mathrm{528}\:\mathrm{000} \\ $$$$ \\ $$$$\:\:\:{Things}\:{are}\:{very}\:{clear}\:{in}\:{front}\:{of}\:{you} \\ $$$$\:{and}\:{so}\:{far}\:{i}\:{have}\:{not}\:{felt}\:{that}\:{i}'{ve}\:{overlooked} \\ $$$$\:{one}\:{of}\:{the}\:{cases} \\ $$$$ \\ $$

Commented by Acem last updated on 04/Nov/22

A) C_( 3) ^( 5) C_( 2) ^( 2) 2!= 20 B) (C_( 4) ^( 6) C_( 2) ^( 2) + C_( 3) ^( 6) C_( 3) ^( 3) ) 2! = 70 not 50 C) Our main question “20 books to 5 stu.” Can you note your method to compare our distribution methods without using special tables

$$\left.{A}\right)\:{C}_{\:\mathrm{3}} ^{\:\mathrm{5}} \:{C}_{\:\mathrm{2}} ^{\:\mathrm{2}} \:\:\mathrm{2}!=\:\mathrm{20} \\ $$$$\left.{B}\right)\:\left({C}_{\:\mathrm{4}} ^{\:\mathrm{6}} \:{C}_{\:\mathrm{2}} ^{\:\mathrm{2}} \:+\:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} \:{C}_{\:\mathrm{3}} ^{\:\mathrm{3}} \right)\:\mathrm{2}!\:=\:\mathrm{70}\:{not}\:\mathrm{50} \\ $$$$ \\ $$$$\left.{C}\right)\:{Our}\:{main}\:{question}\:``\mathrm{20}\:{books}\:{to}\:\mathrm{5}\:{stu}.'' \\ $$$$\:\:{Can}\:{you}\:{note}\:{your}\:{method}\:{to}\:{compare}\:{our} \\ $$$$\:{distribution}\:{methods}\:{without}\:{using} \\ $$$$\:{special}\:{tables} \\ $$

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

if you are not convinced that your result is wrong, why not verify your solution with some simple cases which can be easily checked “manuallly”? for example example 1: 5 books to 2 students, answer should be 20. example 2: 6 books to 2 students, answer should be 50.

$${if}\:{you}\:{are}\:{not}\:{convinced}\:{that}\:{your} \\ $$$${result}\:{is}\:{wrong},\:{why}\:{not}\:{verify}\:{your}\: \\ $$$${solution}\:{with}\:{some}\:{simple}\:{cases}\: \\ $$$${which}\:{can}\:{be}\:{easily}\:{checked}\: \\ $$$$``{manuallly}''?\:{for}\:{example} \\ $$$${example}\:\mathrm{1}:\:\mathrm{5}\:{books}\:{to}\:\mathrm{2}\:{students},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{answer}\:{should}\:{be}\:\mathrm{20}. \\ $$$${example}\:\mathrm{2}:\:\mathrm{6}\:{books}\:{to}\:\mathrm{2}\:{students},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{answer}\:{should}\:{be}\:\mathrm{50}. \\ $$

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

A) 5 books to 2 students: answer is 20. this can be checked: C_2 ^5 ×2=20 i can get the same result via coef. of x^5 in expansion of 5!(e^x −1−x)^2 . B) 6 books to 2 students: answer is 50. this can be checked: C_2 ^6 ×2+C_3 ^6 =15×2+20=50, not 70 ! in fact one can enumerate all these 50 cases easily. i can get the same result via coef. of x^6 in expansion of 6!(e^x −1−x)^2 . C) 20 books to 5 students: the result is the coef. of x^(20) in the expansion of 20!(e^x −1−x)^5 , which is 64 341 775 001 400. i can get this result “manually” in an other way as following. to distribute 20 books into 5 groups such that each group gets at least 2 books, there are following 30 possibilities: 12+2+2+2+2 ⇒((20!)/(12!(2!)^4 4!))×5! 11+3+2+2+2 ⇒((20!)/(11!3!(2!)^3 3!))×5! 10+4+2+2+2 ⇒((20!)/(10!4!(2!)^3 3!))×5! 10+3+3+2+2 ⇒((20!)/(10!(3!2!)^2 2!2!))×5! 9+5+2+2+2 ⇒((20!)/(9!5!(2!)^3 3!))×5! 9+4+3+2+2 ⇒((20!)/(9!4!3!(2!)^2 2!))×5! 9+3+3+3+2 ⇒((20!)/(9!(3!)^3 2!3!))×5! 8+6+2+2+2 ⇒((20!)/(8!6!(2!)^3 3!))×5! 8+5+3+2+2 ⇒((20!)/(8!5!3!(2!)^2 2!))×5! 8+4+4+2+2 ⇒((20!)/(8!(4!2!)^2 2!2!))×5! 8+4+3+3+2 ⇒((20!)/(8!4!2!(3!)^2 2!))×5! 8+3+3+3+3 ⇒((20!)/(8!(3!)^4 4!))×5! 7+7+2+2+2 ⇒((20!)/((7!)^2 (2!)^3 2!3!))×5! 7+6+3+2+2 ⇒((20!)/(7!6!3!(2!)^2 2!))×5! 7+5+4+2+2 ⇒((20!)/(7!5!4!(2!)^2 2!))×5! 7+5+3+3+2 ⇒((20!)/(7!5!2!(3!)^2 2!))×5! 7+4+4+3+2 ⇒((20!)/(7!3!2!(4!)^2 2!))×5! 7+4+3+3+3 ⇒((20!)/(7!4!(3!)^3 3!))×5! 6+6+4+2+2 ⇒((20!)/((6!)^2 4!(2!)^2 2!2!))×5! 6+6+3+3+2 ⇒((20!)/(2!(6!)^2 (3!)^2 2!2!))×5! 6+5+5+2+2 ⇒((20!)/(6!(5!)^2 (2!)^2 2!2!))×5! 6+5+4+3+2 ⇒((20!)/(6!5!4!3!2!))×5! 6+5+3+3+3 ⇒((20!)/(6!5!(3!)^3 3!))×5! 6+4+4+4+2 ⇒((20!)/(6!2!(4!)^3 3!))×5! 6+4+4+3+3 ⇒((20!)/(6!(4!)^2 (3!)^2 2!2!))×5! 5+5+5+3+2 ⇒((20!)/(3!2!(5!)^3 3!))×5! 5+5+4+4+2 ⇒((20!)/(2!(5!)^2 (4!)^2 2!2!))×5! 5+5+4+3+3 ⇒((20!)/(4!(5!)^2 (3!)^2 2!2!))×5! 5+4+4+4+3 ⇒((20!)/(5!3!(4!)^3 3!))×5! 4+4+4+4+4 ⇒((20!)/((4!)^5 5!))×5! Σ: 64 341 775 001 400

$$\left.{A}\right)\:\mathrm{5}\:{books}\:{to}\:\mathrm{2}\:{students}:\: \\ $$$${answer}\:{is}\:\mathrm{20}.\:{this}\:{can}\:{be}\:{checked}: \\ $$$${C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{2}=\mathrm{20} \\ $$$${i}\:{can}\:{get}\:{the}\:{same}\:{result}\:{via}\:{coef}.\:{of} \\ $$$${x}^{\mathrm{5}} \:{in}\:{expansion}\:{of}\:\mathrm{5}!\left({e}^{{x}} −\mathrm{1}−{x}\right)^{\mathrm{2}} . \\ $$$$ \\ $$$$\left.{B}\right)\:\mathrm{6}\:{books}\:{to}\:\mathrm{2}\:{students}: \\ $$$${answer}\:{is}\:\mathrm{50}.\:{this}\:{can}\:{be}\:{checked}: \\ $$$${C}_{\mathrm{2}} ^{\mathrm{6}} ×\mathrm{2}+{C}_{\mathrm{3}} ^{\mathrm{6}} =\mathrm{15}×\mathrm{2}+\mathrm{20}=\mathrm{50},\:{not}\:\mathrm{70}\:! \\ $$$${in}\:{fact}\:{one}\:{can}\:{enumerate}\:{all}\:{these} \\ $$$$\mathrm{50}\:{cases}\:{easily}. \\ $$$${i}\:{can}\:{get}\:{the}\:{same}\:{result}\:{via}\:{coef}.\:{of} \\ $$$${x}^{\mathrm{6}} \:{in}\:{expansion}\:{of}\:\mathrm{6}!\left({e}^{{x}} −\mathrm{1}−{x}\right)^{\mathrm{2}} . \\ $$$$ \\ $$$$\left.{C}\right)\:\mathrm{20}\:{books}\:{to}\:\mathrm{5}\:{students}: \\ $$$${the}\:{result}\:{is}\:{the}\:{coef}.\:{of}\:{x}^{\mathrm{20}} \:{in}\:{the} \\ $$$${expansion}\:{of}\:\mathrm{20}!\left({e}^{{x}} −\mathrm{1}−{x}\right)^{\mathrm{5}} ,\:{which} \\ $$$${is}\:\mathrm{64}\:\mathrm{341}\:\mathrm{775}\:\mathrm{001}\:\mathrm{400}. \\ $$$${i}\:{can}\:{get}\:{this}\:{result}\:``{manually}''\:{in}\:{an} \\ $$$${other}\:{way}\:{as}\:{following}. \\ $$$${to}\:{distribute}\:\mathrm{20}\:{books}\:{into}\:\mathrm{5}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{gets}\:{at}\:{least} \\ $$$$\mathrm{2}\:{books},\:{there}\:{are}\:{following}\:\mathrm{30} \\ $$$${possibilities}: \\ $$$$\mathrm{12}+\mathrm{2}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{12}!\left(\mathrm{2}!\right)^{\mathrm{4}} \mathrm{4}!}×\mathrm{5}! \\ $$$$\mathrm{11}+\mathrm{3}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{11}!\mathrm{3}!\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{10}+\mathrm{4}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{10}!\mathrm{4}!\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{10}+\mathrm{3}+\mathrm{3}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{10}!\left(\mathrm{3}!\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{9}+\mathrm{5}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{9}!\mathrm{5}!\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{9}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{9}!\mathrm{4}!\mathrm{3}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{9}+\mathrm{3}+\mathrm{3}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{9}!\left(\mathrm{3}!\right)^{\mathrm{3}} \mathrm{2}!\mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{8}+\mathrm{6}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{8}!\mathrm{6}!\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{8}+\mathrm{5}+\mathrm{3}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{8}!\mathrm{5}!\mathrm{3}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{8}+\mathrm{4}+\mathrm{4}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{8}!\left(\mathrm{4}!\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{8}+\mathrm{4}+\mathrm{3}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{8}!\mathrm{4}!\mathrm{2}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{8}+\mathrm{3}+\mathrm{3}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{8}!\left(\mathrm{3}!\right)^{\mathrm{4}} \mathrm{4}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{7}+\mathrm{2}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\left(\mathrm{7}!\right)^{\mathrm{2}} \left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{2}!\mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{6}+\mathrm{3}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{7}!\mathrm{6}!\mathrm{3}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{5}+\mathrm{4}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{7}!\mathrm{5}!\mathrm{4}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{5}+\mathrm{3}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{7}!\mathrm{5}!\mathrm{2}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{4}+\mathrm{4}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{7}!\mathrm{3}!\mathrm{2}!\left(\mathrm{4}!\right)^{\mathrm{2}} \mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{7}+\mathrm{4}+\mathrm{3}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{7}!\mathrm{4}!\left(\mathrm{3}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{6}+\mathrm{4}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\left(\mathrm{6}!\right)^{\mathrm{2}} \mathrm{4}!\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{6}+\mathrm{3}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{2}!\left(\mathrm{6}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{5}+\mathrm{5}+\mathrm{2}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{6}!\left(\mathrm{5}!\right)^{\mathrm{2}} \left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{6}!\mathrm{5}!\mathrm{4}!\mathrm{3}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{5}+\mathrm{3}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{6}!\mathrm{5}!\left(\mathrm{3}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{4}+\mathrm{4}+\mathrm{4}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{6}!\mathrm{2}!\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{6}+\mathrm{4}+\mathrm{4}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{6}!\left(\mathrm{4}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{5}+\mathrm{5}+\mathrm{5}+\mathrm{3}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{3}!\mathrm{2}!\left(\mathrm{5}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{5}+\mathrm{5}+\mathrm{4}+\mathrm{4}+\mathrm{2}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{2}!\left(\mathrm{5}!\right)^{\mathrm{2}} \left(\mathrm{4}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{5}+\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{4}!\left(\mathrm{5}!\right)^{\mathrm{2}} \left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!\mathrm{2}!}×\mathrm{5}! \\ $$$$\mathrm{5}+\mathrm{4}+\mathrm{4}+\mathrm{4}+\mathrm{3}\:\Rightarrow\frac{\mathrm{20}!}{\mathrm{5}!\mathrm{3}!\left(\mathrm{4}!\right)^{\mathrm{3}} \mathrm{3}!}×\mathrm{5}! \\ $$$$\mathrm{4}+\mathrm{4}+\mathrm{4}+\mathrm{4}+\mathrm{4}\:\Rightarrow\frac{\mathrm{20}!}{\left(\mathrm{4}!\right)^{\mathrm{5}} \mathrm{5}!}×\mathrm{5}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Sigma:\:\mathrm{64}\:\mathrm{341}\:\mathrm{775}\:\mathrm{001}\:\mathrm{400} \\ $$

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

Commented by Acem last updated on 05/Nov/22

Hello and good evening my best friend At first and for the C_2 ^6 ×2+C_3 ^6 =15×2+20=50, not 70 ! you say..... or 3 books for each student C_3 ^6 C_3 ^3 you know as well that we have 2 ways for distrubting cause we have different books that is why i multiply C_3 ^6 by 2 to get finally 70 ways waiting for your reply Sir, and sure i will be back for the important issue about main qusetion Thanks for your efforts

$${Hello}\:{and}\:{good}\:{evening}\:{my}\:{best}\:{friend} \\ $$$${At}\:{first}\:{and}\:{for}\:{the} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{6}} ×\mathrm{2}+{C}_{\mathrm{3}} ^{\mathrm{6}} =\mathrm{15}×\mathrm{2}+\mathrm{20}=\mathrm{50},\:{not}\:\mathrm{70}\:! \\ $$$${you}\:{say}.....\:{or}\:\mathrm{3}\:{books}\:{for}\:{each}\:{student}\:{C}_{\mathrm{3}} ^{\mathrm{6}} {C}_{\mathrm{3}} ^{\mathrm{3}} \\ $$$$\:{you}\:{know}\:{as}\:{well}\:{that}\:{we}\:{have}\:\mathrm{2}\:{ways}\:{for}\:{distrubting} \\ $$$$\:{cause}\:{we}\:{have}\:{different}\:{books}\:{that}\:{is}\:{why} \\ $$$$\:{i}\:{multiply}\:{C}_{\mathrm{3}} ^{\mathrm{6}} \:{by}\:\mathrm{2}\:{to}\:{get}\:{finally}\:\mathrm{70}\:{ways} \\ $$$$\:{waiting}\:{for}\:{your}\:{reply}\:{Sir},\:{and}\:{sure}\:{i}\:{will} \\ $$$$\:{be}\:{back}\:{for}\:{the}\:{important}\:{issue}\:{about}\:{main}\:{qusetion} \\ $$$${Thanks}\:{for}\:{your}\:{efforts} \\ $$

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

as you wrote C_3 ^6 you have already counted all cases. example: say the books are a,b,c,d,e,f. when we select books a,b,c for the first person, the second person gets automactically the books d,e,f. when the first person gets d,e,f, the second persons gets automatically a,b,c. that means both cases are completely included in C_3 ^6 . this is because both persons get equal number of books. it′s different when a person gets 2 books and the other person 4 books, then we must write 2×C_2 ^6 . think about it, should be clear. example 4 books for 2 persons: a+bcd or bcd+a ✓ b+acd or acd+b ✓ c+abd or abd+c ✓ d+abc or abc+d ✓ ab+cd or cd+ab (?!) ac+bd or bd+ac (?!) ad+bc or bc+ad (?!) bc+ad or ad+bc (?!) bd+ac or ac+bd (?!) cd+ab or ab+cd (?!) you can see we can not multiply by 2 for the cases when both people get the same number of books.

$${as}\:{you}\:{wrote}\:{C}_{\mathrm{3}} ^{\mathrm{6}} \:{you}\:{have}\:{already}\: \\ $$$${counted}\:{all}\:{cases}.\:{example}:\:{say}\:{the} \\ $$$${books}\:{are}\:{a},{b},{c},{d},{e},{f}.\:{when}\:{we}\:{select} \\ $$$${books}\:{a},{b},{c}\:{for}\:{the}\:{first}\:{person},\:{the} \\ $$$${second}\:{person}\:{gets}\:{automactically} \\ $$$${the}\:{books}\:{d},{e},{f}.\:{when}\:{the}\:{first}\:{person} \\ $$$${gets}\:{d},{e},{f},\:{the}\:{second}\:{persons}\:{gets} \\ $$$${automatically}\:{a},{b},{c}.\:{that}\:{means}\:{both} \\ $$$${cases}\:{are}\:{completely}\:{included}\:{in}\:{C}_{\mathrm{3}} ^{\mathrm{6}} . \\ $$$${this}\:{is}\:{because}\:{both}\:{persons}\:{get}\: \\ $$$${equal}\:{number}\:{of}\:{books}.\:{it}'{s}\:{different} \\ $$$${when}\:{a}\:{person}\:{gets}\:\mathrm{2}\:{books}\:{and}\:{the} \\ $$$${other}\:{person}\:\mathrm{4}\:{books},\:{then}\:{we}\:{must} \\ $$$${write}\:\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{6}} .\:{think}\:{about}\:{it},\:{should}\:{be} \\ $$$${clear}. \\ $$$${example}\:\mathrm{4}\:{books}\:{for}\:\mathrm{2}\:{persons}: \\ $$$${a}+{bcd}\:{or}\:{bcd}+{a}\:\checkmark \\ $$$${b}+{acd}\:{or}\:{acd}+{b}\:\checkmark \\ $$$${c}+{abd}\:{or}\:{abd}+{c}\:\checkmark \\ $$$${d}+{abc}\:{or}\:{abc}+{d}\:\checkmark \\ $$$${ab}+{cd}\:{or}\:{cd}+{ab}\:\left(?!\right) \\ $$$${ac}+{bd}\:{or}\:{bd}+{ac}\:\left(?!\right) \\ $$$${ad}+{bc}\:{or}\:{bc}+{ad}\:\left(?!\right) \\ $$$${bc}+{ad}\:{or}\:{ad}+{bc}\:\left(?!\right) \\ $$$${bd}+{ac}\:{or}\:{ac}+{bd}\:\left(?!\right) \\ $$$${cd}+{ab}\:{or}\:{ab}+{cd}\:\left(?!\right) \\ $$$${you}\:{can}\:{see}\:{we}\:{can}\:{not}\:{multiply}\:{by}\:\mathrm{2}\: \\ $$$${for}\:{the}\:{cases}\:{when}\:{both}\:{people}\:{get}\:{the} \\ $$$${same}\:{number}\:{of}\:{books}. \\ $$

Commented by Acem last updated on 05/Nov/22

Yes i agree with you, and with this in mind, let′s ask the following essential question: How does the whole process work? A) Mr. w please come to get 3 books, “Books selection process C_( 3) ^( 6) ”here they are. the rest books for Acem...... it′s ok! B) “Books selection process C_( 3) ^( 6) ” {Math, Fluid mechanics, Steam generators} Now me and you like these science fields, the rest books are about other things. and here, i see that there′s still a different, ′in fact i want these books for me not you′ so, don′t we have different ways to distribute? with diferent books and people

$${Yes}\:{i}\:{agree}\:{with}\:{you},\:{and}\:{with}\:{this}\:{in}\:{mind},\:{let}'{s} \\ $$$$\:{ask}\:{the}\:{following}\:{essential}\:{question}: \\ $$$$\:{How}\:{does}\:{the}\:{whole}\:{process}\:{work}? \\ $$$$ \\ $$$$\left.{A}\right)\:{Mr}.\:{w}\:{please}\:{come}\:{to}\:{get}\:\mathrm{3}\:{books},\: \\ $$$$\:\:\:\:``{Books}\:{selection}\:{process}\:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} \:''{here}\:{they}\:{are}. \\ $$$$\:{the}\:{rest}\:{books}\:{for}\:{Acem}......\:{it}'{s}\:{ok}! \\ $$$$\: \\ $$$$\left.{B}\right)\:``{Books}\:{selection}\:{process}\:{C}_{\:\mathrm{3}} ^{\:\mathrm{6}} \:'' \\ $$$$\:\:\:\left\{{Math},\:{Fluid}\:{mechanics},\:{Steam}\:{generators}\right\}\: \\ $$$$\:{Now}\:{me}\:{and}\:{you}\:{like}\:{these}\:{science}\:{fields},\:{the}\:{rest} \\ $$$$\:{books}\:{are}\:{about}\:{other}\:{things}. \\ $$$$\:{and}\:{here},\:{i}\:{see}\:{that}\:{there}'{s}\:{still}\:{a}\:{different}, \\ $$$$\:'{in}\:{fact}\:{i}\:{want}\:{these}\:{books}\:{for}\:{me}\:{not}\:{you}' \\ $$$$\:{so},\:{don}'{t}\:{we}\:{have}\:{different}\:{ways}\:{to}\:{distribute}? \\ $$$$\:{with}\:{diferent}\:{books}\:{and}\:{people} \\ $$

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

i can′t follow you sir. there are 50 ways to distribute 6 different books to 2 students such that each gets at least 2 books. do you agree or not? if you agree, that′s fine, we can close the case. if you don′t agree, you are wrong, and i also close the case, because i can′t contribute any more. you only need to enumerate all possibilities and then you′ll see that there are indeed only 50 different ways.

$${i}\:{can}'{t}\:{follow}\:{you}\:{sir}.\: \\ $$$${there}\:{are}\:\mathrm{50}\:{ways}\:{to}\:{distribute}\:\mathrm{6}\: \\ $$$${different}\:\:{books}\:{to}\:\mathrm{2}\:{students}\:{such}\: \\ $$$${that}\:{each}\:{gets}\:{at}\:{least}\:\mathrm{2}\:{books}. \\ $$$${do}\:{you}\:{agree}\:{or}\:{not}? \\ $$$${if}\:{you}\:{agree},\:{that}'{s}\:{fine},\:{we}\:{can}\:{close} \\ $$$${the}\:{case}. \\ $$$${if}\:{you}\:{don}'{t}\:{agree},\:{you}\:{are}\:{wrong},\:{and} \\ $$$${i}\:{also}\:{close}\:{the}\:{case},\:{because}\:{i}\:{can}'{t} \\ $$$${contribute}\:{any}\:{more}.\:{you}\:{only}\:{need}\:{to}\: \\ $$$${enumerate}\:{all}\:{possibilities}\:{and}\:{then} \\ $$$${you}'{ll}\:{see}\:{that}\:{there}\:{are}\:{indeed}\:{only} \\ $$$$\mathrm{50}\:{different}\:{ways}. \\ $$

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