The problem is simpler to think about if we first imagine that the objects are in a line, not in a circle. Perhaps if I were a harmonic analyst I would find it easier to think in circles.
We analyze the line problem, and then make a simple adjustment to solve the circle problem.
Let $L(n,k)$ be the number of ways of choosing $k$ objects from a line of $n$ objects, no two of them adjacent.
There are two possibilities (i) the first (leftmost) object in the line is among the chosen ones and (ii) the first object is not among the chosen ones.
In case (i), the second object cannot be chosen, which leaves $L(n-2,k-1)$ possibilities for the remaining $k-1$. In case (ii), we need to choose $k$ objects from $n-1$, giving $L(n-1,k)$ possibilities.
We have obtained the recurrence
$$L(n,k)=L(n-2,k-1)+ L(n-1,k).$$
With the obvious initial conditions, the solution of this recurrence is given by
$$L(n,k)=\binom{n-k+1}{k}.$$
It is not hard to check that the above is a solution of the recurrence. There should be a bijective argument that makes it obvious, you choose the people and pad the choice with a blank seat to the right, most of the time, but I can't quite make it work. Annoying! Boring old induction does work, however.
Now let's adapt this to the circle, by bending the line and connecting the ends. Let $C(n,k)$ be the number of allowed choices.
Some of the choices that were counted in $L(n,k)$ now become "bad." They are the arrangements in a line in which we chose both endpoints. Everything else is fine.
Now count the bad line choices. We chose both endpoints, so could not choose the next to end points, so needed to choose $k-2$ points from $n-4$. There are $L(n-4,k-2)$ ways to do that. Putting things together, we get
$$C(n,k)=\binom{n-k+1}{k}-\binom{n-k-1}{k-2}$$
(if $k$ satisfies the obvious constraint).
It was obvious!: For the line, think of the following operation. Imagine
$n-k+1$ chairs laid out in a row, and choose any $k$ of them. Now for every chosen chair except the rightmost one, add a chair immediately to the right of the chosen chair. Then we get an acceptable arrangement.
Conversely, given any acceptable arrangement, remove the chair immediately to the right of every chosen chair except the rightmost one. In this way we obtain a choice of $k$ chairs from $n-k+1$ chairs. It follows that obviously
$$L(n,k)=\binom{n-k+1}{k}$$
and the fussing with a recurrence for $L(n,k)$ was unnecessary. But I am leaving the original solution intact, to show that clean solutions do not come instantly, at least not to me.
The order in which the balls are drawn is irrelevant. Let the green balls have numbers from $1$ to $10$ and the blue balls have numbers from $11$ to $20$.
There are ${20\choose 5}=15\,504$ subsets of five balls from all twenty balls.
There are are ${10\choose 5}=252$ subsets of five balls from these twenty balls, when only the numbers from $1$ to $10$ may occur.
It follows that $15\,504-252={\bf 15\,252}$ subsets of five contain at least one ball with number $>10$.
Best Answer
Here is a copy of an answer I gave to a similar post (I'll edit it to fit your question a bit better):
The formula for permutations is $\frac{n!}{(n-k)!}$
The formula for combinations is $\frac{n!}{k!(n-k)!}$
(Here I assume you understand the permutation formula a bit). To derive the second from the first let's make an example... say we are finding the combinations of $3$ numbers from a set of $4$, say $1,2,3,$ and $4$. If we use the permutation formula we get our answer as $24$, but we can see we over-counted... for example, $1,2,3$ and $1,3,2$ were counted separately. Since we are taking sets of size $3$ and there are $3!$ ways to arrange a set of size $3$, we have over-counted by a factor of $3!$. Generalizing this, if we use the permutation formula to get combinations we will always over-count by $k!$, where $k$ is the size of the set we take from the whole, and hence the formula for combinations above (we say order does not matter with combinations because of that division... every set we take is unique no matter how we arrange it. This is not true for permutations).
EDIT: Since you are new to this, we get the permutations formula by first choosing one from a set of $n$, then $(n-1)$, then $(n-2)$, and we do this until we get $(n-k+1)$, yielding $P(n,k)=n(n-1)(n-2)(n-3)\ldots(n-k+1)$ (The last case avoids $0$ when $n=k$). We then multiply by (n-k)! to get $$P(n,k)=\frac{n(n-1)(n-2)(n-3)\ldots(n-k+1)(n-r)!}{(n-r)!}$$ $$= \frac{[n(n-1)(n-2)\ldots(n-k+1)]\cdot[(n-k)(n-k-1)\ldots(3)(2)(1)]}{((n-r)!}$$ $$= \frac{n!}{(n-r)!}$$ (The last step requires an observation about the descending nature of the numerator... I leave it as a math puzzle to you!)