How many ways are there to select 4 people in a round circular table from 15 people sitting on it such that no two people selected are adjacent? I want to know what is wrong with my approach.
I approached  the problem like this:

The red circles,  are the positions of different people, and the green lines suggest a random selection which agrees with the condition. (ie green lines are selected people).

$x_i$ denotes the gap between two consecutive selections. For the selections to be non consecutive $x_i \geq 1$. There is one - one correspondence between increasing/decreasing the size of the gap and the selections of the people on the table.
counting the number of gaps: it should be $15-5+1=11$.
so now $ x_1 + x_2 + x_3 + x_4 = 11$ and $x_i \geq 1 $, so we just have to find the number of integral solutions for this, and it would correspond to the number of selection.
The above condition is equivalent to the number of integral solutions of:
$ y_1 + y_2 + y_3 + y_4 = 7$, where $y_i \geq 0$
so our answer should be $ 10 \choose 3$ $=120$, but the answer should be $450$, what am i doing wrong?
 A: It's a bit hard to follow your diagrams.
Sticking with what I think is your approach:  There are $4$ gaps formed by your four choices, and the lengths of those gaps must sum to $11$.  There are $\binom {10}3=120$ ways to populate those gaps with positive natural numbers.  There are $15$ choices for the "first" selection.  But of course any choice out of the $4$ could have been the "first" one so you are counting each selection $4$ times.  Thus the desired result is $$\frac {\binom {10}3\times 15}{4}=450$$
Here's an example to illustrate the idea behind the calculation.  Suppose the gap sequence was $\{3,3,3,2\}$.  Those sum to $11$ so it's a valid choice.  That alone does not give us a choice of four people though, you need to start somewhere.  Say $1$ was the first selection.  Then use that gap sequence to get the selection $(1,5,9,13)$.  However you could have also obtained that selection by starting with $5$ and using the gap sequence $\{3,3,2,3\}$ or by starting with $9$ and using $\{3,2,3,3\}$ or by starting with $13$ and using $\{2,3,3,3\}$.  Thus this procedure generates each selection exactly four times, so you must divide by $4$ to correct.
A: You need a reference point.  Suppose it is the north end of the table.  Then, if we proceed clockwise around the table from the north end of the table, there are three possibilities:

*

*one of the selected people sits at the north end of the table

*one of the selected people sits to the immediate right of the person at the north end of the table

*neither of these two seats is occupied

Notice that by using a reference point, we have transformed the circular problem into a linear problem.
One of the selected people sits at the north end of the table:  In this case, there are four gaps, one to the right of each selected person as we proceed clockwise around the table.  Let $x_i$ be the number of people seated in each gap.  Then the requirement that no two people are adjacent means each $x_i$ is a positive integer.  Since four of the fifteen people are selected, $11$ are not.  Hence,
$$x_1 + x_2 + x_3 + x_4 = 11 \tag{1}$$
is an equation in the positive integers.  The number of solutions of equation $1$ in the positive integers is
$$\binom{11 - 1}{4 - 1} = \binom{10}{3}$$
One of the selected people sits to the immediate right of the person at the north end of the table:  In this case, there are four gaps, one to the left of each selected person as we proceed clockwise around the table.  By symmetry, there are
$$\binom{11 - 1}{4 - 1} = \binom{10}{3} = 120$$
such selections.
Neither the seat at the north end of the table nor the seat to its immediate right is occupied:  In this case, there are five blocks of people who are not selected, the people before the first selected person and one block to the right of each selected person as we proceed clockwise around the table.  Let $x_0$ be the number of people before the first selected person.  Let $x_i$, $1 \leq i \leq  3$, be the number of people between the $i$th selected person and the $(i + 1)$st selected person.  Let $x_4$ be the number of people to the right of the fourth selected person.  Since neither the seat the north end of the table nor the seat to its immediate right is occupied, each $x_i$ is a positive integer.  Hence,
$$x_0 + x_1 + x_2 + x_3 + x_4 = 11 \tag{2}$$
is an equation in the positive integers with
$$\binom{11 - 1}{5 - 1} = \binom{10}{4}$$
solutions.
Total:  Since the three cases considered above are mutually exclusive and exhaustive, the number of ways four people can be selected from $15$ people seated at a circular table so that no two of the selected people sit in adjacent seats is
$$2\binom{10}{3} + \binom{10}{4} = 2 \cdot 120 + 210 = 450$$
