Math Is Fun Forum

Hi;

I got 945 (as Bobby did), also by playing "Spot the Pattern".

OEIS gives that answer too, and the formula (2*n)!/(n!*2^n). In our case n is the number of pairs of athletes.

The answer is the product of the first n odd numbers: 1*3*5*7*9 = 945.

I did the first four elements of the pattern by hand, but stopped there to play STP as the fifth was going to mean I'd be writing into the middle of next week!

Another way of saying it:

OE, EA & AO are invalid arrangements, each numbering 5! = 120.

If OE is immediately preceded by A (ie, AOE) or immediately followed by A (ie, OEA) - it can't be both - that reduces AO's or EA's total (respectively) by 4!, viz: 5!-4! = 120-24 = 96.

Similarly, this scenario for OE also applies to EA & AO (which therefore also number 96), and so we have 3*96 = 288 invalid arrangements.

Total number of valid arrangements = 720 possibles-288 invalids = 6!-3(5!-4!) = 432.

I forget exactly how I arrived at my formula in post #6, but I think the method is wrong, even though it gives the correct answer.

I'm now getting this:

Hi koushikghosh19, and welcome to the forum!

Your dead penguin count is incorrect, as it doesn't include babies:

Works well!

I like that way, filtering out invalids.

I got the permutations like you did, but then got stuck trying to select valids.

...ones that require the wisdom of Solomon and ones that don't.

Then they can't do this:

Hi Bobby;

You got Me on that one!

Penguin Puzzle Phormula: in the case that halPh the babies are Phemale (and ignoring the Partial Penguin conundrum), we could use a compound interest Phormula, covered by MIPH here. It works.

Kermit the Phrog and Phozzie Bear - Phootloose and Phancy-Phree

Provokingly put.

Properly processed, partial-penguin portions' participation produces precise population poll projection.

Hi josh22312, and welcome to the forum!

Your solution is correct. It is one of two solutions to the puzzle with its current revised wording, which was a failed attempt in June 2012 to produce a single-solution version after it was discovered that the original wording had multiple solutions.

Puzzles like this typically have just one solution, and the fact that there are more has attracted quite a bit of discussion on the forum in several threads.

Here is a link to the most recent discussion. It also contains the latest revision (see below), to which there is only one solution. It was first posted 3 Feb 2013, here, but hasn't made its way to MIF's puzzle page.

  1. The person in the middle watches Desperate Housewives
  2. Bob is 46
  3. The person who watches the Simpsons is next to the person who lives in a youth hostel
  4. The person going to Africa is behind Rachael
  5. The person who lives in a village is 52
  6. The person who is going to Australia has straight hair
  7. The person travelling to Africa watches Desperate Housewives
  8. The 14-year-old is at the end of the queue
  9. Amy watches Eastenders
10. Eilish doesn't live on a farm
11. The person heading to Italy has long hair
12. Keeley lives in a village
13. The 46-year-old is bald
14. The fourth in the queue is going to England
15. The people who watch Desperate Housewives & Neighbours are standing next to each other
16. The person who watches Coronation Street stands next to the person with an afro
17. The 21-year-old lives in a youth hostel
18. The person who watches Corrie has curly hair
19. The 81-year-old isn't going to England
20. The person who is travelling to France lives in a town
21. A person next to Rachael has an afro
22. The person who watches Neighbours has long hair

Maybe you'd like to try it!

Very yummy lunch, thanks, followed late afternoon by even yummier birthday party food!

That's a neat trick, using %n to choose specific output lines! It does away with introducing another variable, unlike the following variations of your code...

Mine:

p=Select[Range[10000],PrimeQ[#^2+27]&&PrimeQ[#+3]&];
Total[p]
Length[p]

SE1:

p=Select[Range[10000],PrimeQ[#^2+27]&&PrimeQ[#+3]&];#[p]&/@{Total,Length}

SE2:

p=Select[Range[10000],PrimeQ[#^2+27]&&PrimeQ[#+3]&];Through[{Total,Length}[p]]