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3blue1brown
3blue1brown

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Revisiting Moser's circle problem

One of the earliest videos on this channel was about Moser's circle problem, a famous puzzle where the answers seem to follow a certain pattern, but one which turns out to be misleading.

I thought it might be nice to re-record the video, since the sound quality and pacing of early videos were really not great, and the problem has so many nice lessons within it. Also, posting the "How they fool ya" song, and a few shorts relevant to this sort of popped it back up in relevance.

As I dove in, I ended up changing quite a bit, adding meaningful new sections and animations. It really is a fun puzzle, enjoy!

Revisiting Moser's circle problem

Comments

What a lovely gem: both the content and the presentation.

Edith Dubiner

The division by 2 is just part of the computation for (n choose 2). The relation to Pascal's triangle is that our final formula looks like 1 + (n choose 2) + (n choose 4), which can be interpreted as adding the 0th, 2nd, and 4th elements of the nth row in the triangle. Or by looking one row up, and considering how the triangle is defined, it's the same as adding the first 5 elements of the (n-1)th row. The point about half the triangle was just in the case n = 10 when adding the first 5 elements of the 9th row happens to be exactly half of the row.

3blue1brown

noice, so why is it only requiring just one half* of the pascal triangle, is it because of the divide by 2 factor of the number of permutations? in other words if your question didnt ask for uniqueness of the intersecting line's directions then it would perfectly match pascals triangle?


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