Menu Close

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




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

Leave a Reply

Your email address will not be published. Required fields are marked *