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?

Another try: Consider the 11 positions of the white dots around the circle. There is no possibility of arranging them so that a rotation repeats the pattern, because 11 is a prime (any shuffling of dots must exchange pairs, or rotate groups of 3, or...). So for each linear order of the dots (there are $\binom{24}{13}$ of those) can be rotated into 24 positions, giving again $\frac{1}{24} \binom{24}{13}$. There are two ways of flipping the logo over: Through two dots, or through the lines connecting adyacent dots. But the last one will have an odd number of white dots at one side and an even one at the other, can't give the same. By exchanging left and right through vertically oposite dots, by the same reasoning one has to be red and the other white. This leaves 6 red and 5 white dots to be arranged the same way at each side, and with the selection of red or white dot at the top this then gives $2 \binom{11}{6}$ configurations that can't be distinguised by flipping over. So, allowing flipover the total number of logos is (half of the original have to be subtracted as they result from flipping over, but the ones which are the same flipped over have to be added back in):
$$
\frac{1}{2} \cdot \cfrac{1}{24} \cdot \binom{24}{13} + 2 \binom{11}{6}
$$
