Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$.
We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$s and 5 $B$s, thereby showing that there are ${10\choose 5}$ such sets.
Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$-th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$s and 9 $B$s. As no two numbers were consecutive, we know that in our string no two $A$s are consecutive. We can now remove exactly one $B$ from between each pair of $A$s to get a string with 5 $A$s and 5 $B$s. And clearly this is a bijection, as from each string with 5 $A$s and 5 $B$s we can reconstruct one original set by reversing the construction.
Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$, and the answer is $1750 \bmod 1000 = \boxed{750}$.