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.

So, my incorrect answer

\[ \sum_{k = 0}^{\lfloor n/p \rfloor} \frac{n!}{p^k} \]

should be corrected to

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

Okay, that matches what you said.

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

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.

So, my incorrect answer

\[ \sum_{k = 0}^{\lfloor n/p \rfloor} \frac{n!}{p^k} \]

should be corrected to

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

Okay, that matches what you said.