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 2021-01-03 

Christmas (2020) is over

Showing all comments about the post Christmas (2020) is over. To return to the blog post, click here.

Comments

Comments in green were written by me. Comments in blue were not written by me.
Dec 15th was my favorite. I kept making a logical error and had to restart, so it took me way to long, but I really enjoyed it.

The 16th was just cheeky, after I spent way to much time on the 15th it was nice to have something like that!

It's been 16+ years since I did any probability or combinations and permutations, so it was nice to brush off that part of my brain, not that I did any of them well, but it should serve me well when my kids start doing them and ask me for help.
Dave
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You can solve the Dec 21 puzzle using the principle of inclusion/exclusion:

-There are 6! total ways of arranging 6 numbers.
-Now we have to exclude the ones that don't fit. How many ways have 2 following 1? You can think of 12 as a pair, so you're arranging 12/3/4/5/6 in any order, so there are 5! ways to do this. And there are (5 choose 1)=5 total pairs that might exist, so there are 5*5! ways that have either 12, 23, 34, 45, or 56.
-Of course, we've double counted some that have more than one pair. (This is where inclusion/exclusion comes in, we have to include them back in). So how many have, say, 12 and 45? Well now we're arranging 12/3/45/6, so there are 4! ways to do so. There are (5 choose 2)=10 different pairs, so the double counting was 10*4!.
-We continue this on, and inclusion/exclusion says we keep alternating adding and subtracting as we add more pairs, so the answer is:
6!
- (5 choose 1) * 5!
+ (5 choose 2) * 4!
- (5 choose 3) * 3!
+ (5 choose 4) * 2!
- (5 choose 5) * 1!
= 309
Todd
×6   ×4   ×3   ×3   ×3     Reply
There seems to be a missing diagram for the answer to the Dec 2 puzzle.
Kai
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@(anonymous): Thanks, links corrected
Matthew
                 Reply
Messed up HTML. What I mean is sequence A000255 on OEIS.
(anonymous)
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Happy New Year, one and all!

I think your links point to last year’s calendar puzzles. As for how to solve the puzzle on 21 December, - permutations of [1,...,n+1] having no substring [k,k+1] - is an option.
(anonymous)
                 Reply
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