Number of ways to select $K$ people from $N$ people such that at least $r$ people are there in between them Number of ways to select $K$ people from $N$ people (who are aligned in a line)  such that at least $r$ people are there in between them
Ex: I want to select 3 people from 10 people (who are aligned in a line) such that at least 2 people should be there between the selected people 
May I know the resources for learning this type of problems , also what is the solution if people are arranged in circular manner
 A: Without the distance constraint, the answer would be $\binom{n}k$. Selections of people correspond to sequences of $k$ ones and $n-k$ zeroes. 
To account for this new constraint, we need to ensure that there are at least $r$ zeroes in each of the $k-1$ gaps between the ones. So, set aside $r(k-1)$ zeroes, leaving $k$ ones and $n-k-r(k-1)$ zeroes. Arrange these remaining numbers in a line in $\binom{n-r(k-1)}{k}$ ways. Finally, insert $r$ zeroes into each of the $k-1$ regions between ones. This produces a selection where any two ones have at least $r$ zeroes between them, and every such selection is produced uniquely. Therefore, the number of ways is
$$
\binom{n-r(k-1)}{k}.
$$
Edit: For the circular variant, 


*

*Set aside $kr$ zeroes, and $1$ one. 

*Arrange the remaining numbers in a line, in $\binom{n-rk-1}{k-1}$ ways

*Place the one at the beginning of the line.

*Place $r$ zeroes after each of the $k$ ones.

*Arrange this list clockwise around the circle, in $n$ ways. 
This seems to give a result of $n\binom{n-rk-1}{k-1}$. However, this procedure gives special treatment to the initally removed one, so there is overcounting by a factor of $k$. Therefore, the actual count is
$$
\frac{n}k\binom{n-rk-1}{k-1}
$$
