1. Consider all the subsets of $\{1,2,3, \ldots, 2018,2019\}$ having exactly 100 elements. For each subset, take the greatest element. Find the average of all these greatest elements.
1. Consider all the subsets of $\{1,2,3, \ldots, 2018,2019\}$ having exactly 100 elements. For each subset, take the greatest element. Find the average of all these greatest elements.
Solution 1: Let $M$ be the average that we are computing. First, there are $\binom{2019}{100}$ ways to choose a 100 -element subset. Next, if $x$ is the largest element, then $x \geq 100$, and there are $\binom{x-1}{99}$ subsets having $x$ as the largest element. Hence
$$
M=\frac{\sum_{x=100}^{2019} x\binom{x-1}{99}}{\binom{2019}{100}} \text {. }
$$

But note that
$$
x\binom{x-1}{99}=100\binom{x}{100} .
$$

Using this fact and the hockey stick identity, we have
$$
\begin{aligned}
M & =\frac{100 \sum_{x=100}^{2019}\binom{x}{100}}{\binom{2019}{100}} \\
& =\frac{100\binom{2020}{101}}{\binom{2019}{100}} \\
& =\frac{100 \cdot 2020}{101} \\
& =2000
\end{aligned}
$$

Solution 2: As in Solution 1, the required average $M$ can be written as
$$
\begin{aligned}
M & =\frac{\sum_{x=100}^{2019} x\binom{x-1}{99}}{\binom{2019}{100}} \\
& =\frac{100\binom{99}{99}+101\binom{100}{99}+\cdots+2019\binom{2018}{99}}{\binom{2019}{100}} .
\end{aligned}
$$

We can simplify this expression by grouping the terms in the numerator in specific ways, applying the hockey stick identity to each group of terms, and then applying the hockey stick identity again to the resulting terms. This can be done in several ways, two of which are shown next:

Way 2.1: Note that
$$
\begin{aligned}
\binom{2019}{100} M= & 100\binom{99}{99}+101\binom{100}{99}+\cdots+2019\binom{2018}{99} \\
= & 2020\left[\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2018}{99}\right]-\left[1920\binom{99}{99}+1919\binom{100}{99}+\cdots+\binom{2018}{99}\right. \\
= & 2020\left[\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2018}{99}\right] \\
& -\left[\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2018}{99}\right] \\
& -\left[\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2017}{99}\right] \\
& -\cdots-\binom{99}{99} \\
= & 2020\binom{2019}{100}-\binom{2019}{100}-\binom{2018}{100}-\cdots-\binom{100}{100} \\
= & 2020\binom{2019}{100}-\binom{2020}{101} .
\end{aligned}
$$
where the last two lines follow from the hockey stick identity.
Hence, $M=2020-\frac{\binom{2020}{101}}{\binom{2019}{100}}=2020-\frac{2020}{101}=2000$.
Way 2.2: Let
$$
X:=\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2018}{99}=\binom{2019}{100}
$$

Then
$$
\begin{aligned}
\binom{2019}{100} M= & 100\binom{99}{99}+101\binom{100}{99}+\cdots+2019\binom{2018}{99} \\
= & 100\left[\binom{99}{99}+\binom{100}{99}+\cdots+\binom{2018}{99}\right] \\
& +\left[\binom{100}{99}+\cdots+\binom{2018}{99}\right] \\
& +\left[\binom{101}{99}+\cdots+\binom{2018}{99}\right] \\
& +\cdots+\binom{2018}{99} \\
= & 100 X+\left[X-\binom{100}{100}\right]+\left[X-\binom{101}{100}\right]+\cdots+\left[X-\binom{2018}{100}\right]
\end{aligned}
$$

Thus,
$$
\begin{aligned}
M X & =100 X+1919 X-\left[\binom{100}{100}+\binom{101}{100}+\cdots+\binom{2018}{100}\right] \\
& =2019 X-\binom{2019}{101}
\end{aligned}
$$
which gives
$$
M=2019-\frac{\binom{2019}{101}}{\binom{2019}{100}}=2020-\frac{2020}{101}=2000 .
$$