Math Is Fun Forum

Hi;

I suppose that the answer could also be...

Alhough Calligar's puzzle differs from the film's, the following may still apply because of the "based on" connection:
Kaito says about his puzzle, "What is this? Kindergarten?" (English dubbed version)/"Can it get any simpler?" (English subtitles version), and so my second answer may well be the solution as it is the easiest of my three options. But then, most puzzles are child's play to him, and so maybe my most intricate answer – the first – wins.

EDIT: I think the second and third answers are unlikely solutions as both can be solved without using any numbers from the puzzle. The first answer uses them all, but can be shortened by omitting step 2 from the explanation as it only serves to identify an interesting relationship that doesn't seem to be a solution requirement.

Sorry, Anna, but I only have a very basic understanding of permutations, which is mainly from research I've done since you started posting permutation problems. I wasn't taught permutations at school.

anonimnystefy and others here can help you with the formulas you need.

Anyway, for what it's worth, here's how I calculated the first-choice v2 and v3 and all the second-choice varieties:

I went with what I know and like to use - an Excel spreadsheet, even though I knew it would be the long way round compared to using permutation formulas (which I did try to write, btw, but your problem had too many tricks for my basic knowledge, and my research didn't unearth the info I was looking for). 

I got Mathematica to list all 3125 (ie, 5^5) permutations (with repeat digits) of 5-digit numbers comprising the digits 1 to 5 and copied the list into an Excel spreadsheet, which I used for the rest of the project. There I whittled the list down to 2220 by eliminating permutations that contained the same digit more than twice.

I then got Excel to count the permutations that had distinct digits of specified lengths for first-choice questions:
(a) 5 distinct digits for v1 (no repeat digits)............of which there were 120    )
(b) 4 distinct digits for v2 (1 pair of repeat digits)...of which there were 1200   )  total 2220
(c) 3 distinct digits for v3 (1 pair of repeat digits)....of which there were 900    )

I could simply have deducted 120 and 1200 from 2220 to get 900, but I used the Excel count to verify that this component was going well. 

Directly underneath the row of 2220 first-choice permutations I created a row of corresponding second-choice permutations, with each one formed from leftover question numbers after taking into account those that were used by first-choice permutations. I got Excel to display them with their digits in ascending order to help with eliminating duplicates.

And that's all there was to that!

Hi Anna;

Please let me know if these three jump sequences - expressed in {1,2,-1} terms - enable Skippy to reach the bowl:

A. 1,1,1,1,1,1,1,-1,1
B. 2,1,1,2,1,-1,1
C. 2,2,1,2,-1,1

Hi Nehushtan;

Please see post #33.

Maybe also the description of my method in post #34. It's not mathematical, but may help understand the logic I used...which basically was to list all permutations of jumps (including invalid ones, as I found it easier to account for the permutations this way), and to then deduct from that list all invalid permutations (those being the backwards first jumps and the backwards last jumps).

Hi Bobby,

I can't explain my method better than graphically, so here goes, in terms of {1,2,-1}:

Happy for someone to point out the flaw in my logic!

Gotta have some lunch...catch you later.

What do you mean?

I like Nehushtan's logic, but he left something out in both answers and included something in the first that gave the correct answer there.

How did you do it? Do you think all your answers are valid?

Hi Anna;

I'm going to try to repost this, as the page wouldn't load. Maybe listing all 2040 permutations was the problem. So here goes, with less this time:

Here are the first 50 permutations from the total of 2040.

The permutations are in ascending order of concatenated strings of chosen questions (as per the post #14 method), with each student's picks also in ascending order. These two features enabled me to eliminate all duplicates.

eg, {12 12 34 35 54} isn't listed, as it's a duplicate of {12 12 34 35 45} (ie, student E picks questions {4,5}, in any order). Only the first version of a set that contains such duplicates is listed.   

  #  A  B  C  D  E
  1  12 12 34 35 45
  2  12 12 34 45 35
  3  12 12 35 34 45
  4  12 12 35 45 34
  5  12 12 45 34 35
  6  12 12 45 35 34
  7  12 13 23 45 45
  8  12 13 24 35 45
  9  12 13 24 45 35
 10  12 13 25 34 45
 11  12 13 25 45 34
 12  12 13 34 25 45
 13  12 13 34 45 25
 14  12 13 35 24 45
 15  12 13 35 45 24
 16  12 13 45 23 45
 17  12 13 45 24 35
 18  12 13 45 25 34
 19  12 13 45 34 25
 20  12 13 45 35 24
 21  12 13 45 45 23
 22  12 14 23 35 45
 23  12 14 23 45 35
 24  12 14 24 35 35
 25  12 14 25 34 35
 26  12 14 25 35 34
 27  12 14 34 25 35
 28  12 14 34 35 25
 29  12 14 35 23 45
 30  12 14 35 24 35
 31  12 14 35 25 34
 32  12 14 35 34 25
 33  12 14 35 35 24
 34  12 14 35 45 23
 35  12 14 45 23 35
 36  12 14 45 35 23
 37  12 15 23 34 45
 38  12 15 23 45 34
 39  12 15 24 34 35
 40  12 15 24 35 34
 41  12 15 25 34 34
 42  12 15 34 23 45
 43  12 15 34 24 35
 44  12 15 34 25 34
 45  12 15 34 34 25
 46  12 15 34 35 24
 47  12 15 34 45 23
 48  12 15 35 24 34
 49  12 15 35 34 24
 50  12 15 45 23 34

Done on Excel spreadsheet. I still haven't been able to write a mathematical formula for it.

Anyway, I'm feeling fairly confident about this result.

Hmmm...a huge number of duplicates slipped under my guard.

I've eliminated them all now (I think!) by using a strengthened version of the post #14 method, and got 2040 as the answer.