And, for the sake of the peanut gallery, this problem reduces to 3 simple problems:

1. How many ways to choose 2 from a set of 5;
2. How many ways to choose 2 from a set of 24;
3. How many ways to choose 2 from a set of 9.

or

5P2 * 24P2 * 9P2 = 794,880

 

I've noticed a lot of members here use computer simulation to check their designs.

I use my designs to check the computer...

If your script had given a different answer, I would have told you there is something wrong with your computer/script.

I actually wasn't using the computer to check my design, but rather preparing as strong a case as I could for when studiot says that the number is actually half of that. The first two times I ran the script I did get a different answer and I knew something was wrong with the script and tracked it down and fixed it. I would have started questioning the design only after I was very confident that the script was right and, even then, I would not have been satisfied with either until both agreed and I fully understood where the discrepancy came from and that it had been resolved correctly.

 

And, for the sake of the peanut gallery, this problem reduces to 3 simple problems:

1. How many ways to choose 2 from a set of 5;
2. How many ways to choose 2 from a set of 24;
3. How many ways to choose 2 from a set of 9.

or

5P2 * 24P2 * 9P2 = 794,880

I completely agree, but that isn't going to satisfy someone that insists that the order matters because they will just maintain that your solution doesn't apply to this problem.

 

Order doesn't matter in this problem because there are five vowels for four character positions. If there were six positions (e.g., a vowel in the first and sixth positions), then order becomes more important.

Hey, where is strantor?

 

Order doesn't matter in this problem because there are five vowels for four character positions. If there were six positions (e.g., a vowel in the first and sixth positions), then order becomes more important.

Hey, where is strantor?

Nope. Still wouldn't matter.

If you have six positions and each position must be a letter with first and last being vowels and no repeats anywhere, then the number of possible combinations is:

(5)(4)(24)(23)(22)(21) = 5,100,480

Again, let's work a toy problem that can be enumerated easily.

We have four positions, first and last must be vowels and we have three vowels available along with two consonants, so let's use {aeixy} as our alphabet. My claim is that the number of legal combinations is:

(3)(2)(3)(2) = 36

aexi eaxi iaxe
aeyi eayi iaye
aixe eixa iexa
aiye eiya ieya
axei exai ixae
axie exia ixea
axye exya ixya
axyi exyi ixye
ayei eyai iyae
ayie eyia iyea
ayxe eyxa iyxa
ayxi eyxi iyxe

So, are there any legal combinations not in this list? Any illegal combinations that are in the list? Any duplications in the list?

 

If we ignore the numerals at the end

Has anyone noticed that the set of all passwords generatable by

VLLV

is different from the set of all passwords generatable by

VVLL

Where V = vowel and L = any letter

 

If we ignore the numerals at the end

Has anyone noticed that the set of all passwords generatable by

VLLV

is different from the set of all passwords generatable by

VVLL

Where V = vowel and L = any letter

Naturally, by definition. But the number of permutations is the same.

 

If we ignore the numerals at the end

Has anyone noticed that the set of all passwords generatable by

VLLV

is different from the set of all passwords generatable by

VVLL

Where V = vowel and L = any letter

Hint: Take the solution set for VLLV, and for each solution in the set transpose the first L with the last V. This will produce the a new set of permutations equivalent to VVLL.

 

Order doesn't matter in this problem because there are five vowels for four character positions. If there were six positions (e.g., a vowel in the first and sixth positions), then order becomes more important.

What WBahn said. You can solve the problem in any order you wish, as long as you recognize that any combination with 4 vowels in position 2 through 5 is not part of the solution -- by definition.

 

If we ignore the numerals at the end

Has anyone noticed that the set of all passwords generatable by

VLLV

is different from the set of all passwords generatable by

VVLL

Where V = vowel and L = any letter

Sure, the set is different, but the number of legal combinations is the same.

Enumerate a toy example and see for yourself.

 

...answer tomorrow.

Tomorrow will be over in 3 hours, and I've got to get some sleep. Any chance on getting the real answer before then?

 

I'm waiting for a report from the setting board to see if there's a piece missed in the question since their 'official' answer is

5*24*23*2*9*8 = 397440

 

I'm waiting for a report from the setting board to see if there's a piece missed in the question since their 'official' answer is

5*24*23*2*9*8 = 397440

Why do you require a report, or an "official" answer? You must have your own answer, and evidence to back it up. Come on, we did a lot of work here. How about you?

 

Studiot,

It took you 34 minutes to reject my original answer.

Now, show your answer and your work.

 

Joey and WBahn, you have an error in the fourth place.

So what is this error that we have in the fourth place?

I maintain that what I have done is consistent with the rules as I interpreted them in Post #8, namely:

Assumption #1: "Letter" means one of the 26 characters of the case-insensitive English alphabet.
Assumption #2: "Vowel" means one of the 5 characters from the set {a,e,i,o,u}.
Assumption #3: "no repetition of characters is allowed" means that any given character can be used at most once within the entire password (as opposed to just meaning that the same character can't be used twice in succession).

So please either tell me how I have misinterpreted the rules or where my solution is inconsistent with that interpretation.

 

I'm waiting for a report from the setting board to see if there's a piece missed in the question since their 'official' answer is

5*24*23*2*9*8 = 397440

Think about a set of rules that would be consistent with this equation:

The first letter must be a vowel.
The second letter can be any letter except the vowel used in the first position and one other vowel that will be set aside and not used.
The third letter can be any letter except one of the first two letters or the other vowel that was previously set aside.
The fourth letter can be any vowel except the first letter, the vowel that was previously set aside, or a second vowel that will now be set aside and not used.
The fifth and sixth letters can be any numeric digit greater than 0 that are not repeated.

Now, doesn't this seem a bit more restrictive than the actual rules? Wouldn't it therefore make sense that it would have a lower number of possibilities than the actual rules.

 

I've read your rules for the fourth character several times, and it isn't consistent with the rules as stated originally. Why are you restricting using the vowel that was set aside? Wasn't it set side to be used here??? And why are you setting aside a vowel THAT WON'T BE USED??? And why are you changing the rules?

 

And why are you changing the rules?

Nobody is intentionally changing the rules.

I posted the question as I originally saw it, except that I added the bit about not being case sensitive and removed the alleged exam board's answer.

I later posted an interpretation of it, but I may have missed something and I'm still not sure if what I originally posted was the exact wording as used by the exam board.

Since exam boards are very very rarely wrong in their published answers I am trying to obtain an exact copy of the question.

 

Studiot, have you worked out the problem yourself without relying on the board's answer? It took me three attempts, but I arrived at the same answer as WBahn and joeyd99, using a slightly different approach.

 

I understand WBahn's answer and note that joey's working is equivalent,

But I also note that there is a question as to how the password is to be formed.

If you allow the selection of the second vowel before the first consonant WBahn and joey are correct.

If however you insist that the characterers are chosen in order of appearance in the password then their analysis is flawed since there are 25 characters available at the instant of choice of the second character.
This fact is not reflected in joey's permutation expression.

And yes, I have played with a few smaller models to see how things work out, but I am reserving judgement until I have the full information