💾 Archived View for bbs.geminispace.org › u › stack › 2160 captured on 2024-05-26 at 17:21:50. Gemini links have been rewritten to link to archived content
⬅️ Previous capture (2024-05-10)
-=-=-=-=-=-=-
Re: "I was thinking about how many unique games can be had in..."
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.
2023-06-17 · 11 months ago
🚀 mbays · 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.
🚀 stack [mod] · 2023-06-20 at 13:58:
@mbays: you are totally right, I completely missed the local permutations...
🍀 gritty [OP] · 2023-06-20 at 23:45:
@mbays nice. thanks for the explanation
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.