将$r$个不同东西分给$n+p$人使其中某$n$人每人至少分到1个

hbghlyj

George Chrystal - Algebra_ An elementary textbook, Vol 2 Page 33 Exercise Ⅲ. 27
The number of ways in which $r$ different things cen be distributed among $n+p$ persons so that certain $n$ of those persons may each have one at least is
$$
g_r=(n+p)^r-n(n+p-1)^r+\frac{n(n-1)}{2 !}(n+p-2)^r-\ldots
$$
Hence prove thet
$$
S_1=S_2=\ldots = S_{n-1}=0, \quad S_n=n !, \quad S_{n+1}=\left(\frac{n}{2}+p\right)(n+1) ! .
$$
(Wolstenholme.)

kuing

英文太多没细看，但当看到那串 S 的结果时，我想起了这帖：
forum.php?mod=redirect&goto=findpost& … d=6540&pid=33651
不知有没有联系。

色k

我好像看懂了那个 $g_r$ 用的是容斥原理？