How many combinations of coloured dots (with restrictions)? My friend is designing a logo. The logo can essentially be reduced to 24 coloured dots arranged in a circle, and they can be either red or white. We want to produce a individual variation of this logo for each employee. That, if I have worked it out right, (since this appears analogous to a 24-bit binary string), means we could have an individual logo for 2^24 employees, obviously way more than we need.
But of course, we don't really want logos that don't have a lot of white dots as they may look too sparse. So we stipulate that there must always be at least half + 1 = 13 in the logo. How many combinations does that restrict us to?
My initial thought is 12 (half) + 1 + 2^11, but I'm not good enough to prove it.
Also, how can we generalise this formula for $x$ dots, $y$ individual colours and at least $n$ colours of a single type? If that's too general, what about just the case $y = 2$ as we have above?

If rotations of the circle are allowed, you need to apply Pólya's coloring theorem. The relevant group for just rotations of 24 elements is $C_{24}$,
whose cycle index is:
$$
\zeta_{C_{24}}(x_1, \ldots x_{24}) = \frac{1}{24} \sum_{d \mid 24} \phi(d)x_d^{24 / d}
 = \frac{1}{24} \left( x_1^{24} + x_2^{12} + 2 x_3^{8} + 2 x_4^{6} + 3 x_6^4 + 4 x_8^3 + 6 x_{12}^2 + 8 x_{24} \right)
$$
For 13 red and 11 white ones (use $r$ and $w$ for them) you want the coefficient of $r^{13} w^{11}$ in $\zeta_{C_{24}}(r + w, r^2 + w^2, \ldots, r^{24} + z^{24})$. The only term that can provide exponents 13 and 11 is the first one:
$$
[r^{13} w^{11}] \zeta_{C_{24}}(r + w, r^2 + w^2, \ldots, r^{24} + z^{24})
= [r^{13} w^{11}] \frac{1}{24} (r + w)^{24}
= \frac{1}{24} \binom{24}{13}
$$
Flipping over is left as an excercise ;-)
(I'm sure that as soon as I post this, somebody will post a simple reason why this is so by considering that 24 is even, and 13 and 11 odd...).
