Problem: 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 . Solution: The answer is $750$. Constructing these sets is quite burdensome. Recall that if $A \backslash\{x ,y\in A: x-y \ge \ell\} = \{ x,y \in A: x-y <\ell \}$. We will use this principle to see that it is much easier to count the 5-element subsets of the first $14$ natural numbers that have no consecutive elements. That is, we replace $\ell$ with $2$ in the above. At first, it may seem that counting jumps is extremely hard to do, but here is a very nice way to interpret it. Notice that there is a bijection between subsets of an arbitrary subset $A \subseteq \mathbb{N}$ and the strings comprised of $Ys and $Ns; Consider $\{ 1,2,3 \}$. In this simple case, we can specify each subset by putting a $N$ to say "No, don't include this element" and $Y$ to say "Yes, do include this element." The idea is that, for instance, $NYN$ corresponds to $\{2\}$ and $YYN$ corresponds to $\{1,2\}$. Our problem now becomes, how many arrangements of five $Ys and nine $Ns are there such that there are two consecutive $Ys. We can construct them! Consider writing $\_Y\_Y\_Y\_Y\_Y\_$. Now, we need to be sure that in the four spaces between the five $Ys there is a $N$, so we just put them there from the start. After this, we have five $Ns left over to distribute among six spots, giving a total of $\binom{5 + 6-1}{6 - 1} = \binom{10}{5}$ such strings. The answer is thus $\binom{14}{5} - \binom{10}{5} = 1750$ and so $M = 750$.