26.Oct.22

I'll just keep using my email address as the example here, since it's where the idea came from, it's my website, and it's good branding maybe? Anyway, without dots, my email address is ADAMAWRIGHT1985

The problem assumes we can't start or end the address with a dot, and we can't have multiple dots between characters.

That means every gap between two characters could have a dot there. There's only two possibilities, then:

1. 1. a gap WILL have a dot, or

   2. a gap WILL NOT have a dot

(Anything that has exactly two options is called a binary choice. Knowing this term doesn't give us an advantage in this problem, but it's nice to learn.)

We also need to know how many gaps there actually are. We can think of it like this:

If I have a log, and I consider that log to currently be uncut (we're going to ignore the fact that it came from a tree that had to be cut down and pruned), then I have one continuous piece of wood and have made zero cuts.

If I then cut that log parallel to its round ends, I will have two continuous pieces of wood and have made one cut.

If I cut one of the pieces parallel to its round ends, I will have three continuous pieces of wood and have made two cuts.

Notice that the number of pieces of wood (P) is always one more than the number of cuts (C): P = C + 1

We can think of the caps as cuts and the letters as pieces. So, if I have 15 letters, then 15 = C + 1 will give us 14 = C, or 14 gaps.

Each gap may either contain a dot or not contain a dot. So, let's say the first gap has a dot in it. The next gap could either have a dot or no dot, so that's still two possibilities. And, if the first gap does not have a dot in it, the second gap could again either have a dot or no dot, so that's two more possibilities. Each of these four possibilities meets with the possibilities of the third gap, which either does have a dot or does not, making eight possibilities.

Each time, the number of possibilities is doubling. Thus, the number of possibilities is going to be two times itself a number of times equal to the number of gaps: 2¹⁴ = 16,384 possible email addresses.