Let $A$ be the set of all $4$ digit numbers $a_1a_2a_3a_4$ such that $a_1 < a_2 < a_3 < a_4$, then what is $n(A)$ equal to? How can you solve this problem relatively quickly using combinatorics? I found it really interesting.
Let $A$ be the set of all $4$ digit numbers $a_1a_2a_3a_4$ such that $a_1 < a_2 < a_3 < a_4$, then what is $n(A)$ equal to?

$7,6,5,4,3,2$ in second positions  give $6,5,4,3,2,1$ cases respectively of the  trees below in which for $7$ one has one case, for $6$ three, for $5$ six, for $4$ ten, for $3$ fifteen, and for $2$ twenty one cases (these are the number of endpoints  in each tree). This gives finally 
$$6\cdot1+5\cdot3+4\cdot6+3\cdot10+2\cdot15+1\cdot21=\color{red}{126\text{ cases }}$$

