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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.

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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
×1                 Reply
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
×3   ×1              Reply
There seems to be a missing diagram for the answer to the Dec 2 puzzle.
Kai
                 Reply
@(anonymous): Thanks, links corrected
Matthew
                 Reply
Messed up HTML. What I mean is sequence A000255 on OEIS.
(anonymous)
   ×2              Reply
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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