Problem: How many subsets of $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ have the property that no two of its elements differ by more than 5 ? For example, count the sets $\{3\},\{2,5,7\}$, and $\{5,6,7,8,9\}$ but not the set $\{1,3,5,7\}$. Solution: As problem solvers, we want to be able to control the subsets in some way so they are easier to count. This naturally leads us to partition subsets satisfying this property by the minimum element in this set. Why is the minimum so strong? Well, once we choose one, say $3$, we can only include the elements $\{4,5,6,7,8\}$ in the subset. For each min less than or equal to $7$, we have $2^5$ subsets with that minimum (because we make a binary choice whether or not to include the following $5$ elements in our subset). These mins contribute $7 \cdot 2^5$ to our sum. Then, if the min is $8$, there are $2^4$ subsets satisfying the property; if the min is $7$, there are $2^3$ subsets satisfying the property; ... ; if the min is $12$, there is $2^0 = 1$ subset satisfying this property. We can't forget the empty set, which technically satisfies the condition. In the end, we get $ 7\cdot 2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 + 1 = 256. $