Find the number of ways to select three distinct digits from the set {0,1,2,...,9} if no two consecutive digits can be selected. My answer to this is $8$ choose $3$ because of the non consecutive numbers, the $10$ choices go down to $8$ and we have to pick three digits.
 A: Imagine that you’ve picked $3$ of the digits, no two of them consecutive. Call them $a,b$, and $c$, with $a<b<c$. The remaining $7$ digits have to be distributed amongst four spaces: before $a$, between $a$ and $b$; between $b$ and $c$; and after $c$. Because $a$ and $b$ aren’t consecutive, at least one of the $7$ must go between $a$ and $b$. Similarly, at least one must go between $b$ and $c$. That leaves $5$ that can go in any of the four slots. The question now boils down to this: in how many different ways can we distribute $5$ indistinguishable objects to $4$ slots?

Example: If we put one object in the first slot, two in the second, two in the third, and none in the fourth, we have $$\bullet\;a\,\color{crimson}{\bullet}\,\bullet\,\bullet\;b\,\color{crimson}{\bullet}\,\bullet\,\bullet\,c\;,$$ where the red objects are the required ones, and the black objects are the ones that we were free to insert anywhere. This corresponds to choosing $a=1,b=5$, and $c=9$.

Counting these distributions is a standard stars and bars problem; you’ll find the appropriate formula, with a decent explanation/proof, at the link.
