15. Let $S=\{1,2,3, \cdots, 65\}$. Find the number of 3-element subsets $\left\{a_{1}, a_{2}, a_{3}\right\}$ of $S$ such that $a_{i} \leq a_{i+1}-(i+2)$ for $i=1,2$.
15. Answer. 34220
Solution. Note that the condition $a_{i} \leq a_{i+1}-(i+2)$ for $i=1,2$ is equivalent to that
$$
a_{1}+3 \leq a_{2}, \quad a_{2}+4 \leq a_{3} .
$$

Let $A$ be the set of all 3 -element subsets $\left\{a_{1}, a_{2}, a_{3}\right\}$ of $S$ such that $a_{1}+3 \leq a_{2}$ and $a_{2}+4 \leq a_{3}$.
Let $B$ be the set of all 3 -element subsets $\left\{b_{1}, b_{2}, b_{3}\right\}$ of the set $\{1,2, \cdots, 60\}$.
We shall show that $|A|=|B|=\binom{60}{3}=34220$ by showing that the mapping $\phi$ below is a bijection from $A$ to $B$ :
$$
\phi:\left\{a_{1}, a_{2}, a_{3}\right\} \rightarrow\left\{a_{1}, a_{2}-2, a_{3}-5\right\} .
$$

First, since $\left\{a_{1}, a_{2}, a_{3}\right\} \in A$, we have $a_{1}+3 \leq a_{2}$ and $a_{2}+4 \leq a_{3}$, and so $a_{1}<a_{2}-2<a_{3}-5$, implying that $\left\{a_{1}, a_{2}-2, a_{3}-5\right\} \in B$.
It is clear that $\phi$ is injective,
It is also surjective, as for any $\left\{b_{1}, b_{2}, b_{3}\right\} \in B$ with $b_{1}<b_{2}<b_{3}$, we have $\left\{b_{1}, b_{2}+2, b_{3}+5\right\} \in$ $A$ and
$$
\phi:\left\{b_{1}, b_{2}+2, b_{3}+5\right\} \longrightarrow\left\{b_{1}, b_{2}, b_{3}\right\} .
$$

Hence $\phi$ is a bijection and $|A|=|B|=34220$.