Question

24 distinct books in 4 shelves how many arrangements are possible ?

Answer 1

Answer:

27! / 3!  =  27! / 6

Step-by-step explanation:

Each arrangement corresponds to a permutation of the 24 books and a sum a+b+c+d=24, with a,b,c,d ≥ 0 (allowing for empty shelves), so the first a books in the permuation go on shelf 1, the next b books on shelf 2, etc.

The number of solutions of a+b+c+d=24 with a,b,c,d ≥ 0

= the number of solutions of A+B+C+D=28 with A,B,C,D ≥ 1

= the number of ways of choosing 3 spaces from the 27 spaces between 28 "1"s to put a "+" sign

= $\displaystyle\binom{27}{3}$

The number of permutations of the 24 books is 24!

So all together, the number of arrangements on the shelves (allowing for empty shelves) is

$\displaystyle\binom{27}{3}\times 24!\\\\=\frac{27!}{3!\,24!}\times 24!\\\\=\frac{27!}{3!}$