Help with Permutation Problem (Grade 11 )

[CasualKilla]

Hi there everybody. Helping my brother to prepare for his Math Exam tomorrow and this problem really got me stumped. I could solve it with a tree, but that would take very very long and doesn't seen reasonable.

 

Problem: Create a password that contains 6 characters. The first four characters must be a letter the alphabet, however the first and forth character must be a vowel. The last 2 characters must be a number (1 through 9). How many possible passwords are there. (no repetition of characters)

 

Answer on memo is very confusing: 5*24*23*2*9*8

[fiveworlds]

1.01809221188169112561E+191364

[CasualKilla]

1.01809221188169112561E+191364

 

How did you get to your answer, remember the is a Grade 11 (age 17) level question and its only 3 marks, so I'm guessing the method must be rather simple.

[studiot]

Note your memo answer is the product of 6 factors, and you want a 6 character password.

Is this a coincidence?

 

The first character is a vowel.

There are 5 vowels.

 

So how many ways can this be chosen?

 

So work through the first, second third etc asking each time the same question.

[fiveworlds]

5*24*23*2*9*8

Okay so there is 5 vowels at the start. 26 letters in the alphabet and 9 numbers no letters can be reused.

So there is always a 5 at the start. Then take 1 from 26 because that letter can't be used again. Then it should be 26-1,26-2,4,9,8= 5^25^24^4^9^8. The only thing that is hard to guess is that 4 because if the two before it are vowels it could be a 2 but there is no way to tell.

Edited November 9, 2014 by fiveworlds

[studiot]

Casualkilla do you wish to continue, despite the incorrect interruptions?

[fiveworlds]

Not really incorrect

 

(5^21^20^4^9^8)+(5^21^4^3^9^8)+(5^4^20^3^9^8)+(5^4^3^2^9^8)

Edited November 9, 2014 by fiveworlds

[CasualKilla]

5*24*23*2*9*8

Okay so there is 5 vowels at the start. 26 letters in the alphabet and 9 numbers no letters can be reused.

So there is always a 5 at the start. Then take 1 from 26 because that letter can't be used again. Then it should be 26-1,26-2,4,9,8= 5^25^24^4^9^8. The only thing that is hard to guess is that 4 because if the two before it are vowels it could be a 2 but there is no way to tell.

I believe you can choose a vowel by "accident" when choosing a letter, so its not as simple as 5^25^24^4^9^8, I already thought of that. Is it safe t say the memo is wrong? Its only 3 marks so it implies a simple answer.

[studiot]

No I have already said the memo is not wrong.

Fiveworlds has already said he is just guessing.

 

Have you tried following my method?

Edited November 9, 2014 by studiot

[CasualKilla]

Casualkilla do you wish to continue, despite the incorrect interruptions?

Yes indeed, I'm looking for a explanation, just posting the answer is not helping me anyway.

[studiot]

OK So

 

Position(1) is a vowel and can therefore be chosen in 5 ways.

 

Position (2) can be a vowel but cannot be the vowel in position (1) or the vowel in position (4) (although we don't yet know the second vowel)

 

That means there are 2 (not 1) unavailable letters for position (2).

 

That means there are 26 - 2 = 24 available letters.

 

Can you now complete?

Edited November 9, 2014 by studiot

[CasualKilla]

No I have already said the memo is not wrong.

Fiveworlds has already said he is just guessing.

 

Have you tried following my method?

Yes, I did.

OK So

 

Position(1) is a vowel and can therfore be chosen in 5 ways.

 

Position (2) can be a vowel but cannot be the vowel in position (1) or the vowel in position (4) (although we don't yet know the second vowel)

 

That means there are 2 (not 1) unavailable letters for position (2).

 

That means there are 26 - 2 = 24 available letters.

 

Can you now complete?

No, the 2 still gets me. Im thinking in term of probability tree, but more like possibility trees if you know what I mean. Using your logic, I don't understand why u say p (4) is unavailable, by that same logic why is p (3) also not unavailable.

Edited November 9, 2014 by CasualKilla

[studiot]

 

No, the 2 still gets me.

 

Did you mean the 2 at position (4) or the reduction of 2 for position (2)?

[CasualKilla]

 

Did you mean the 2 at position (4) or the reduction of 2 for position (2)?

I meant the 2 at position (4), sorry I knew that was gonna be ambiguous. But I cant fully grasp the reasoning behind reduction of 2 for position (2) either.

[fiveworlds]

I believe you can choose a vowel by "accident" when choosing a letter, so its not as simple as 5^25^24^4^9^8, I already thought of that. Is it safe t say the memo is wrong? Its only 3 marks so it implies a simple answer.

 

You're right it isn't as simple as that because we have to take each possibility individually like so:

 

Part 1

Find the number permutations without vowels

(5^21^20^4^9^8)

 

Part 2

Find the number of permutations with 1 vowel

(5^21^4^3^9^8)

(5^4^20^3^9^8)

 

Part 3

Find the number of permutations with 2 vowels

(5^4^3^2^9^8)

 

Part 4

Add all of them together

(5^21^20^4^9^8)+(5^21^4^3^9^8)+(5^4^20^3^9^8)+(5^4^3^2^9^8)

Edited November 9, 2014 by fiveworlds

[studiot]

No problem.

 

I don't recommend using the abbreviation p(4) for position 4 since p(something) is normally used for probability(something).

 

Think of your tree (No I don't wnat to draw one either)

 

Position (1) has 5 branches

Let us say postion (1) is an 'a'

 

This must immediately strike other branches that contain an 'a'.

 

So you have a smaller tree than if 'a' was still available.

 

It is exactly the same for whatever vowel goes into position (4) - it must make this vowel unavailable to other branches by striking them out.

It's just that we don't know which one, but it doesn't matter since all the branches have the same pattern, there must just be one fewer for every other letter position.

 

So we have (26 - 2) = 24 in the second position.

 

In the third position we have already struck out the branches that are used by position (4) so we don't need to do it again.

 

So we simply take one off the available total (it could still be a vowel)

 

24 - 1 = 23

 

position (4) the second vowel now comes after 3 preceeding characters have been chosen and these three could all be vowels, leaving 5 - 3 = 2 available vowels.

 

I asume you are OK with the 9 and the 8 for positions (5) and (6) ?

[CasualKilla]

You're right it isn't as simple as that because we have to take each possibility individually like so:

 

Part 1

Find the number permutations without vowels

(5^21^20^4^9^8)

 

Part 2

Find the number of permutations with 1 vowel

(5^21^4^3^9^8)

(5^4^20^3^9^8)

 

Part 3

Find the number of permutations with 2 vowels

(5^4^3^2^9^8)

 

Part 4

Add all of them together

(5^21^20^4^9^8)+(5^21^4^3^9^8)+(5^4^20^3^9^8)+(5^4^3^2^9^8)

Nice simple way to look at it, easier to understand, but takes a bit longer to implement than studiots method, that should be quite easy to explain to the brosef.

 

@Studiot, yes I understand position (8) and (9), they are not linked to the letter sets so you just tack those on.

Studiot I must thank you for that well worded explanation, but maby I'm just stupid because I its not clicking for me.

 

1) Why do we remove the postion (4) vowel branches but not the position (3) letter branches.

2) There is enough vowels to fill all characters, so I don't see why we need to save a spot.

3) I drew a smaller tree with less character options and noticed that the branches where I pick a vowel in the letter slot have less branches when it comes to the vowel branches.

4) You say position (3) and (4) could have been vowels, I agree. But isn't setting it as 2 assuming they where vowels, its not clicking with me yet.

[imatfaal]

I must admit I am still a little foxed.

 

Firstly I cannot see how the order of restriction can make a difference to the overall password strength - thus I would reorder to vowel,vowel, letter, letter, number, number. This easily groks as 5*4*24*23*9*8. Which is not on the memo

 

Secondly. Fiveworlds - you should be using x or * to show multiplication not ^ . that sign is use to show "to the power of" - and your confusion has shown in the size of your first answer. Also your your second perm for using 1 vowel as a letter is slightly wrong (5^4^20^3^9^8) - this should be 5*4*21*3*9*8 as you have not used any consonants before the third character. Again the position cannot make a difference so it should be the same as the other one.

 

And finally - if you take Fiveworlds sum of permutations (corrected as above) and 5*4*24*23*9*8 and evaluate you get to 794880 in both instances.

[studiot]

 

and evaluate you get to 794880

 

Which is exactly double the correct answer.

 

 

4) You say position (3) and (4) could have been vowels, I agree. But isn't setting it as 2 assuming they where vowels, its not clicking with me yet.

 

 

Not quite.

 

positions (2) and (3) could be vowels or consonants.

position (4) is specified as a vowel.

 

imatfaal

What would happen if you only had an alphabet with two vowels, but otherwise the same question?

it is because the fourth one is specified as a vowel that you cannot reorder.

 

 

thus I would reorder to vowel,vowel, letter, letter, number, number

 

By this reordering you can separate the 6 character word into three independent segments.

But you cannot do this when the final letter is specified as a vowel.

 

In your reordering you definitely have 4 vowels available to choose from in position (2)

In the original you may have 4 or 3 or 2

 

This difference must be reflected in different probabilities for the second specified vowel.

[MonDie]

I need to block this forum as soon as I can again.

 

---

 

Analogous dilemma:

You can choose any number 1-5 for space A, but only odd numbers [1,3,5] for space B.

This is small enough to draw a tree diagram, and it comes out to 12 possibilities regardless which space precedes.

 

I suppose you could retain the original order and instead weight it according to the probability of selecting a vowel, but I'm too tired to see whether it would work.

Edited November 10, 2014 by MonDie

[John]

imatfaal's reasoning is correct. A good rule of thumb with these sorts of problems is to start with the most restricted positions, which in this case are the first and fourth.

 

We know the first character must be a vowel, thus there are five possibilities. This leaves four possibilities for the fourth character. Regardless of which two vowels occupy these spaces, we then have 24 and 23 possible characters for the second and third positions, respectively. The fifth and sixth characters can, of course, be treated separately, as everyone here seems to have done.

 

Thus the number of valid passwords is 5(4)(24)(23)(9)(8), as indicated.

 

To work it out more laboriously, we can try to go through in sequence, and we get the following:

VVVVNN: 5(4)(3)(2)(9)(8)
VCVVNN: 5(21)(4)(3)(9)(8)
VVCVNN: 5(4)(21)(3)(9)(8)
VCCVNN: 5(21)(20)(4)(9)(8)

5(4)(3)(2)(9)(8) + 5(21)(4)(3)(9)(8) + 5(4)(21)(3)(9)(8) + 5(21)(20)(4)(9)(8) = 794880 = 5(4)(24)(23)(9)(8)

Edit: I somehow didn't really notice that the latter process is exactly what fiveworlds and imatfaal mentioned in their posts. Blame it on a long day at work.

Edit 2: Another way to see how and why the first method works is to mentally picture the set of all valid passwords. Separate them into blocks depending on what vowels are in the first and fourth positions. So there will be 5(4) blocks in total (one for A__E__, one for A__I__, etc.). For each of these, there are 24(23) letters left to occupy the second and third positions. This is the reasoning encoded by the product 5(4)(24)(23)(9)(8).

Edited November 10, 2014 by John

[fiveworlds]

Edit: I somehow didn't really notice that the latter process is exactly what fiveworlds and imatfaal mentioned in their posts. Blame it on a long day at work.

 

Or it being 1 in the morning here.

[mathematic]

I haven't looked at all the answers, so it is possible someone got it right already. However there are so many wrong so I am putting in my calculation. Start with the two vowels, then the other letters, and finally the digits.

 

5*4*24*23*9*8.

[studiot]

@CasualKilla

 

How did your brother's exam go?

 

I would appreciate it if you posted the exact question as written by the exam board or whoever since there may be something in their wording we have missed, that leads to the solution you say they sent in a memo?

 

Textbooks are rarely wrong in their published answers,

Exam boards even more rarely wrong.

Edited November 12, 2014 by studiot

[overtone]

Textbooks are rarely wrong in their published answers,

Exam boards even more rarely wrong.

Maybe somebody used a choice formula for the vowels (five choose two), instead of a permutation formula, overlooking the importance of order in a password - unless the OP is missing something in the original problem.

 

In my stints as an instructor I developed the suspicion that math text publishers use their customers - which in practice means the students themselves - to proof their problems. I used to give praise and credit for catching errors and ambiguities - trying to find opportunity in what is too often frustrating and angering bafflement - and work as many of them as I had time for myself, to anticipate trouble.

Edited December 1, 2014 by overtone