Let $\mathbb{S}$ be a set of integers, $\max(\mathbb{S})$ be the largest element in $\mathbb{S}$, and $\mid\mathbb{S}\mid$ be the number of elements in $\mathbb{S}$. Find the number of non-empty set $\mathbb{S}\in\{1,2,\cdots,10\}$ satisfying $\max(\mathbb{S})\le\mid\mathbb{S}\mid + 2$.
Let $k=\mid\mathbb{S}\mid$. When $k \ge 9$, the requirements will be automatically satisfied. For each value in $1\le k\le 8$, the elements in corresponding $\mathbb{S}$ can only be selected from $1$, $2$, $\cdots$, $k+2$. There are $\binom{k+2}{k}$ different selections. It follows that the desired answer is $$\binom{10}{10} + \binom{10}{9} + \sum_{k=1}^{8}\binom{k+2}{k} = 11 + \sum_{k=1}^{8}\binom{k+2}{2}= 11 + \left(\binom{11}{3}-\binom{2}{2}\right)=\boxed{175}$$
The second last step utilizes the hockey stick identity.