Re: [理工] [離散] 遞迴 98北大資工
※ 引述《annheilong (方格子)》之銘言：
: D_n = (n-1)(D_n-1 + D_n-2), n>=3, D_1=0, D_2=1
亂序禁位
假設{1,2...n}個數
考慮位置1和其中一個數x
假設x放在位置1，那整數1就有剩下n-1個可能
而剩下的n-2個做亂序為Dn-2
所以(n-1)Dn-2
再來考慮x原本是在位置1
則x扣掉位置1有n-1種可能
剩下的n-1做亂序為Dn-1
所以(n-1)Dn-1
Dn = (n-1)(Dn-1 + Dn-2)
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