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Comment by ๐Ÿš€ stack

Re: "I was thinking about how many unique games can be had in..."

In: s/SpellBinding

And if you don't care about real words, the number of possible unique permutations is 7 * 26 * 25 * 24 * 23 * 22 * 21 * 20

First letter is any of 26. Second is one of remaining 25 (since no duplicates are allowed). Likewise, the third is one of 24, etc. Finally, there are 7 possible games for each set as each of the 7 letters can go in the middle.

๐Ÿš€ stack [mod]

2023-06-17 ยท 1 year ago

3 Later Comments โ†“

๐Ÿš€ mbays ยท 2023-06-20 at 05:58:

@stack You forgot to account for the fact that the order of the outer letters doesn't matter. It should be 7*(26 choose 7) = 7*26!/(19!*7!) = (26*25*24*23*22*21*20)/(6*5*4*3*2*1) = 4604600. To explain: there are 26*25*24*23*22*21*20 ways to fill the spellbinding hexagon with 7 distinct letters, but any permutation of the outer 6 letters gives the same game, and there are 6! = 6*5*4*3*2*1 such permutations.

Interesting that you clamp the possible total scores! I ended up doing that in Zaubuchstabier after getting annoyed at ridiculously high-scoring games, but I'd assumed this was due to the more generative nature of German.

๐Ÿš€ stack [mod] ยท 2023-06-20 at 13:58:

@mbays: you are totally right, I completely missed the local permutations...

๐Ÿ€ gritty [OP] ยท 2023-06-20 at 23:45:

@mbays nice. thanks for the explanation

Original Post

๐ŸŒ’ s/SpellBinding

I was thinking about how many unique games can be had in spellbinding. I've been out of comp sci theory for quite a while but I came up with: 26*6^20 but I'd love someone better at this than me to correct it. 26 center letters * 20 (19?) remaining letters swapped out for each letter on the circle (6). I feel like factorials are involved but maybe not.

๐Ÿ’ฌ gritty ยท 6 comments ยท 2023-06-17 ยท 1 year ago