Using Sticks and Stones for Counting number of Ways

Question

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.

Answer 1

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 $\binom{15}{6}$.

Answer 2

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 $$ \binom{15}{6}= 5005 $$