Warning You need to know some basic group theory terminology to appreciate(I hope you do) the following content. Burnside’s lemma Here is the statement of the Burnside’s lemma: Let $G$ be a group that actions on a set $X$. Denote $X^{g}$ the set of fixed points of $g$ i.e. $\{x \in X | g \cdot x = x\}$, then the number of orbits of the action is equal to $\dfrac{1}{|G|}\sum\limits_{g \in G} |X^g|$....
Problem Let $I_{m} = \int^{2\pi}_{0} \cos(x)\cos(2x)\dots\cos(mx) dx$. For $m$ in $1, 2, \dots, 10$, for which $m$ is $I_m \neq 0$ ? Solution $I_m$ is non-zero if and only if $m \equiv 0, 3 \pmod{4}$. Main idea The official(seemingly? It is in the putnam problem book and a few solutions I found online do the same.) solution is to substitute $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ followed by grouping the terms into $\cos x \cos (2x) \dots \cos (mx) = e^{\text{something}}$, and analyze the something....
Counting binary trees I was thinking about of the problem of balancing a binary tree, and my mind stumbled across to the question “How many different binary trees with labelled nodes can you make without changing the traversal order?”. After figuring out the answer myself I realized that the numbers of such binary trees are just Catalan numebrs (I was not really into combinatorics - now I am). For example, the following has traversal order 1->2->3->4->5->6->7...