Okay, I get it. The multinomial coefficient

${n \choose p, p, \ldots, p, 1, 1, \dots , 1} = \frac{n!}{(p!)^k (1!)^{n-kp}} = \frac{n!}{(p!)^k}$

counts the number of ways to take an \$$n\$$-element set and partition it into _labeled_ parts, \$$k\$$ of size \$$p\$$ and \$$n-kp\$$ of size \$$1\$$. But we don't want _labeled_ parts. We need to ignore the labelings. So we need to divide by \$$k!\$$ and \$$(n-kp)!\$$, to account for permutations of labelings.

$\sum_{k = 0}^{\lfloor n/p \rfloor} \frac{n!}{p^k}$
$\sum_{k = 0}^{\lfloor n/p \rfloor} \frac{n!}{p^k k! (n-kp)!}$