TITLE: Using Sticks and Stones for Counting number of Ways
QUESTION [5 upvotes]: From the first twenty positive integers, how many ways can we select 6 integers so that no two integers from the six chosen ones are consecutive?
I tried using sticks and stones, but my thought process kept running out of gas and I can't see to get on track.

REPLY [5 votes]: There will be $14$ integers not chosen. Write down $14$ occurrences of $\times$, like this:
$$\times \quad\times\quad\times \quad\times\quad\times \quad\times\quad\times \quad\times\quad\times \quad\times\quad\times \quad\times\quad\times \quad\times$$
These determine $15$ gaps, $13$ real ones plus $2$ endgaps. We must choose $6$ of these to slip a Y (yes) into. There is a natural one to one correspondence between the strings of length $20$ made up of $14$ $\times$ and $6$ Y, with no two Y consecutive, and the valid choices. (Number the elements of the string from the left.) The number of ways is therefore $\binom{15}{6}$.

REPLY [2 votes]: The number of ways to select $r$ integers from $(1\ldots n)$ with no two consecutive integers selected is 
$$
\binom{n+1-r}{r}$$
So the answer to your problem is 
$$
\binom{15}{6}= 5005
$$