Marble ring: no two blacks are adjacent Question: How many ways are there to arrange $20$ marbles, $6$ black and $14$ white, in a ring (circular arrangement) such that no two of the black marbles are next to each other? 
Comments: It is easier to find the probability, as there are shortcuts one can make without having to find the number of ways. It is also much easier to do it as if the marbles were in a row. But, in a ring, finding the number of ways is harder. Any ideas? 
 A: To ensure the condition, glue a white marble to the right of each black marble. Now the task is to arrange 6 black(+white) marbles and 8 white marbles. In principle, there are $14\choose 6$ ways to do so.
However, (most) patterns come in sets of 14 differing only by rotation. 
And some (namely $7\choose 3$) point-symmetric patterns come only in sets of 7 differing by rotation.
We should also check for patterns with 7fold or 14fold symmetry, but they can be excluded because there are only 6 black marbles.
In total there are
$$ \frac17{7\choose 3}+\frac1{14}\left({14\choose 6}-{7\choose 3}\right)=217$$
patterns up to rotation (what about reflection?).
A: There are $\binom{8+5}{5}$ partitions of $8$ into $6$ ordered non-negative numbers. Call this set $T$.
There is an action of $\mathbb Z_6$ on $T$ defined by rotation.  Then $T/G$ is the set of orbits under this action. Each orbit must have size $1,2,3,6$.
First, convince yourself that there cannot be any orbits of size $1,2$. 
If there was an orbit of size $1$, then all the elements of your partition would be the same, which isn't possible since $6$ is not a factor of $8$.  
If there was an orbit of size $2$, then there would be a partition of $8$ into three equal numbers.
So there can only be orbits of size $3$ and $6$.
An element of $T$ has orbit size $3$ if it is composed of $2$ identical partitions of $4$ into three non-negative values. There are $\binom{4+2}{2}=15$ such elements of $T$.
There are thus $\binom{8+5}{5}-\binom{4+2}{2}$ elements of $T$ that are in orbits of size $6$.
This means that the number of elements of $T/G$ is:
$$\frac{1}{3}\binom{4+2}{2} + \frac{1}{6}\left(\binom{8+5}{5}-\binom{4+2}{2}\right)$$
More generally, if there are $pa$ white balls and $pb$ black balls, with $p$ prime (or $p=1$), $(a,b)=1$ and $a>b$, then there are:
$$\frac{1}{a}\left((p-1)\binom{a}{b} + \binom{pa}{pb}\right)$$
(This formula is more akin to Hagen's answer, with mine adding some messy $-1$s. But they are the same answer.)
It's more complicated when $p$ is not prime. That's when you need a more complicated inclusion-exclusion argument.
