Bunuel
A shelf contains 20 different books of which 4 are in single volume and the others form sets of 8, 5 and 3 volumes respectively. Number of ways in which the books may be arranged on the shelf, if the volumes of each set are together and in their due order is
(A) \(\frac{20!}{8!5!3!}\)
(B) 7!
(C) 8!
(D) 7*8!
(D) 9!
So what do we have in 20 books : \(A, B, C, D, (E_1, E_2...E_8), (F_1, F_2....F_5), (G_1, G_2, G_3)\)
Now, as all the volumes of a book are together and in due order, we can take \((E_1, E_2...E_8), (F_1, F_2....F_5), (G_1, G_2, G_3)\) as \(E_{(1-8)}, \ \ F_{(1-5)} \ \ G_{(1-3)}\)
Thus we have 7 books : \(A, B, C, D, E_{(1-8)}, \ \ F_{(1-5)} \ \ G_{(1-3)}\). These can be arranged in 7! ways.
Now, due order could mean ascending order or descending order, that is it could be \(E_{(1-8)}, \ \ or \ \ E_{(8-1)}\), so each of sets of E, F and G can be arranged in 2 ways each.
Answer = 7!*2*2*2=8*7!=8!
C