Question Number 101732 by mr W last updated on 04/Jul/20

There are 10 identical mathematics  books, 7 identical physics books  and 5 identical chemistry books.  Find the number of ways to compile   the books under the condition that  same books are not mutually adjacent.

Commented bymr W last updated on 04/Jul/20

if possible please solve generally for  n_m  mathematics books  n_p  physics books  n_c  chemistry books  with n_m ≥n_p ≥n_c

Answered by bemath last updated on 04/Jul/20

case(1) 2×((12!)/(7!.5!))  case(2)9×7×5  totally = ((2×12!)/(7!.5!)) + 9×7×5

Commented bymr W last updated on 04/Jul/20

please explain your answer sir!

Commented bybemath last updated on 04/Jul/20

wrong or correct sir?

Commented bymr W last updated on 04/Jul/20

i have no answer yet.

Commented bybemath last updated on 04/Jul/20

hihihi...

Answered by mr W last updated on 04/Jul/20

in following M is place holder for  a mathematics book, X, Y  are place  holders for physics or chemistry  books.    XYMXMXMXMXMXMXMXMXMXMX  11×2!×((10!)/(6!4!))=4620  XYXMXMXMXMXMXMXMXMXMXM  MXYXMXMXMXMXMXMXMXMXMX  10×(((9!)/(5!4!))+((9!)/(6!3!)))×2=4200  XYMXYMXMXMXMXMXMXMXMXM  MXYMXYMXMXMXMXMXMXMXMX  C_2 ^(10) ×2×2×((8!)/(5!3!))×2=20160  MXYXMXYMXMXMXMXMXMXMXM  9×8×(2×((7!)/(4!3!))+2×((7!)/(5!2!)))=8064  MXYMXYMXYMXMXMXMXMXMXM  C_3 ^9 ×2×2×2×((6!)/(4!2!))=10080    totally:  4620+4200+20160+8064+10080  =47124

Commented bybemath last updated on 04/Jul/20

why sir different to eq 101693

Commented bymr W last updated on 04/Jul/20

they are different! we can′t apply  a general formula for all cases. e.g.  if we had 10 mathematics books,  4 physics books and 4 chemistry books,  there is even no solution.