Rub brand
name suggests is the rub board, we have to rub card, it is necessary to know some common sense, to introduce the following
definition:
Now we have \ (A \) array stores a \ (1-n \) these numbers satisfy the requirements of any one \ (A [I]! = I \) , find how many arrangements
official:
\(f(n)=(n-1)[f(n-1)+f(n-2)]\)
A simple proof:
Set \ (A \) charged \ (B \)
there are two cases
\ (1. \) \ (B \) charged \ (A \) the remaining \ (n-2 \) two articles is \ (F (n-2-) \)
\ (2. \) \ (B \) is not charged \ (A \) then is \ (f (n-1)
\) In addition to discussing \ (A \) outer we may discuss \ (b, c, d ...... \) total \ (n-1 \) case
so \ (f (n) = ( n-1) [f (n-1) + f ( n-2)] \)