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? 

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?).
