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Let $n = p^em$, and let $N$ be the number of subsets of order $p^e$ in a set of order $n$. Determine the congruence class of $N$ modulo $p$.

I think the answer is m and I feel it is a little similar with the prove of Third Sylow Theorem, which shows the number of Sylow p-subgroups is congruent to 1 modulo p by proving only [H] is fixed by H.

Then, could I decompose the N sets into H-orbit and find m sets that fixed by H in this question?

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You're trying to compute $\binom{p^{\large e}m}{p^{\large e}}$ mod $p$. Indeed one can prove $\binom{pa+b}{pc+d}\equiv\binom{a}{c}\binom{b}{d}$ mod $p$ using generating functions mod $p$ (binomial theorem!) and then inducting to prove Lucas' theorem, but we can also invoke orbit-stabilizer for a more combinatorially-pure answer.

To prove a set $X$ has a given size mod $p$, one can equip it with a group action of a $p$-group and then show the number of fixed points of the action has that residue.

The cyclic group $C(p^e)$ acts on itself, and thus acts on $C(p^e)\times\{1,\cdots,m\}$ (leaving the second coordinate fixed). Consider the set $\Lambda$ of all subsets of size $p^e$. Show that the fixed points are precisely the sets $C(p^e)\times\{k\}$ for $k\in\{1,\cdots,m\}$, so there are $m$ fixed points.

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