Math Problem

[Dystopia]

Okay, lets say theres a password.

Lets say the password has 9 characters, and its all contains letters of the alphabet.

Keep into consideration that it could be the same character repeated over and over, for every alphabet.

So like.

* * * * * * * * *

9 characrers.

abcdefghijklmnopqrstuvwxyz
26 characters

If you were to go through ever possibility how many will you have?

I have no clue, I think its 9^26 but Im not exactly sure. That would be 6461081889226673298932241.

If each character had a 1/26 chance of being right, so that is 9^26, that should be right.

What do you guys think?

[Swindler]

dont ask me^^ i suck at math

[bLuE_fLaMe]

ask on yahoo answers, some nerd will know it

[SM-Count]

26^9

[Stress]

You say the letter can be repeated... As far as my low level Algebra can tell the answer is:

29! / (29-6)! = 342014400

Please, can anyone confirm it is so?

[Dystopia]

You say the letter can be repeated... As far as my low level Algebra can tell the answer is:

29! / (29-6)! = 342014400

Please, can anyone confirm it is so?

Whoa! you got a much lower number than I did

[Waisha]

My first thought says it's 26^9. Pretty sure that's correct.

[Azilius]

I tried it with a lower amount of characters and characters in password.


2 letter PW
3 different letters possible

aa
bb
cc
ab
ba
ac
bc
ca
cb

8 options
2^3 = 8


Tried it with 3 and 3 and typed out 27 different answers, couldn't think of any more.

[Dystopia]

I tried it with a lower amount of characters and characters in password.


2 letter PW
3 different letters possible

aa
bb
cc
ab
ba
ac
bc
ca
cb

8 options
2^3 = 8


Tried it with 3 and 3 and typed out 27 different answers, couldn't think of any more.

that makes sense

but with more numbers it should go up right

i mean

aaaaaaaaa
bbbbbbbbb
ccccccccc
....
zzzzzzzzz
abbbbbbbb
acccccccc
...
azzzzzzzz
baaaaaaaa
bbbbbbbbb <- repeated
bcccccccc
.... and it goes on to like
zzzzzzzza
zzzzzzzzb
you know?

o.0?

[Azilius]

To be honest I have no idea

But if it works with lower numbers I don't see why it wouldn't work with the whole alphabet.

[SM-Count]

You have 26 possibilities for the first blank.

You have 26 possibilities for the second blank for each of the possibilities on the first blank. For example, if you chose 'a' for the first blank, you have 26 letters for the second blank, if you chose 'b' you have a separate 26 letters from if you chose 'a'. So we have 26x26 so far, or 26^2.

Stress' factorials are for if you have a limited amount of space for certain items (non-distinct). e.g. 4 spots on a bookshelf for 6 books. 6!/2! or 6!/(6-4)! Distinct would be 6!/4!2! Would not work here because we do replacements, if we couldn't repeat letters then factorials would work.

[fena]

You have 26 possibilities for the first blank.

You have 26 possibilities for the second blank for each of the possibilities on the first blank. For example, if you chose 'a' for the first blank, you have 26 letters for the second blank, if you chose 'b' you have a separate 26 letters from if you chose 'a'. So we have 26x26 so far, or 26^2.

Stress' factorials are for if you have a limited amount of space for certain items (non-distinct). e.g. 4 spots on a bookshelf for 6 books. 6!/2! or 6!/(6-4)! Distinct would be 6!/4!2! Would not work here because we do replacements, if we couldn't repeat letters then factorials would work.

We learned them as permutations versus combinations, but Shadowman is completely right. It'd be 26 to the 9th power.

[Trice]

You have 26 possibilities for the first blank.

You have 26 possibilities for the second blank for each of the possibilities on the first blank. For example, if you chose 'a' for the first blank, you have 26 letters for the second blank, if you chose 'b' you have a separate 26 letters from if you chose 'a'. So we have 26x26 so far, or 26^2.

Stress' factorials are for if you have a limited amount of space for certain items (non-distinct). e.g. 4 spots on a bookshelf for 6 books. 6!/2! or 6!/(6-4)! Distinct would be 6!/4!2! Would not work here because we do replacements, if we couldn't repeat letters then factorials would work.

well explained.
Powers = When you are able to do repeats
Factorials = When you arent able to do repeats

So the answer is 26^9. :]

[LillDev!l]

uhm isnt it 9^26?
26^2 would be like a 26 letter password with a choice of "A" or "B" all the time.
9^26 would be like 9 letter password with 26 possebility's for every time ?

i dont know forgot about the topic a bit

26 nCr 9 also sounds good to me ? since its (what we want) above (total) and since we want a 9 letter password and have 26 letters that would do ?

arg screw it >.< its weekend wont go into it

Good luck on solving it lol^^

P.S. im so bad at explaining things like this in english >.<

[NuclearSilo]

26^9

It's been a long time i didnt do math though

[Inuyasha584]

^ = times? then 9squared times 26? idk guessing..

[NuclearSilo]

^ = power

[Inuyasha584]

oh

[salmissra]

its definitely 26^9 and not 9^26. since there is a 1/26 chance for every letter so for each spot its 1^26 =26...then the second spot is 2^26..so forth...

[No reply]

As we all agree, and the "problem" itself is pretty simple, let's make it a bit more complex:

Let's say you are going to make a password consisting of 3 digits, and 6 letters

- the password has got to start with a letter, and end with a number.
- every letter can only be used once.
- You do of course have caps lock on your keyboard, and is therefor able of using both small an capital letters
- your password should start with a capital letter
- the code cannot contain both the number 8 and 3.


The question is: how many combinations can you
possibly make?


Edit: we're talking about the English alphabet, just to make that clear.

[NuclearSilo]

As we all agree, and the "problem" itself is pretty simple, let's make it a bit more complex:

Let's say you are going to make a password consisting of 3 digits, and 6 letters

- the password has got to start with a letter, and end with a number.
- every letter can only be used once.
- You do of course have caps lock on your keyboard, and is therefor able of using both small an capital letters
- your password should start with a capital letter
- the code cannot contain both the number 8 and 3.


The question is: how many combinations can you
possibly make?


Edit: we're talking about the English alphabet, just to make that clear.

Only Einstein knows the answer

[Simplifique]

As we all agree, and the "problem" itself is pretty simple, let's make it a bit more complex:

Let's say you are going to make a password consisting of 3 digits, and 6 letters

- the password has got to start with a letter, and end with a number.
- every letter can only be used once.
- You do of course have caps lock on your keyboard, and is therefor able of using both small an capital letters
- your password should start with a capital letter
- the code cannot contain both the number 8 and 3.


The question is: how many combinations can you
possibly make?


Edit: we're talking about the English alphabet, just to make that clear.

8.9^13 (you broke my calculator)

[No reply]

8.9^13 (you broke my calculator)

First of all, good morning to whoever lives in a close by timezone. About your answer, Simplifique it is pretty close to what I got : 1,53... * 10^14.
Gonna go trough it once more to check if I'm correct as I was really tired when I made this task in the first place.

Edit: buy a new calculator ( casio CFXb-9850 GC +)
Edit 2: it wasn't as close as I thought.

[NuclearSilo]

explain the formula -_-

[No reply]

26*(52-1)*(51-1)*(50-1)*(49-1)*(48-1)*10*10*10

* ( 7!/5!/2!)

then you just subtract the number of codes which has both 8 and three in it.

[Stress]

Yup. I double checked. Shadowman is right

[SM-Count]

26*(52-1)*(51-1)*(50-1)*(49-1)*(48-1)*10*10*10

* ( 7!/5!/2!)

then you just subtract the number of codes which has both 8 and three in it.

Wouldn't it be 26*(50...)and so on? Since you have to remove both the capital letter & the lower case letter from the first pick.

[Simplifique]

26*(52-1)*(51-1)*(50-1)*(49-1)*(48-1)*10*10*10

* ( 7!/5!/2!)

then you just subtract the number of codes which has both 8 and three in it.

Wouldn't it be 26*(50...)and so on? Since you have to remove both the capital letter & the lower case letter from the first pick.

Its too early in the morning for me to think. Nvm.

[No reply]

Wouldn't it be 26*(50...)and so on? Since you have to remove both the capital letter & the lower case letter from the first pick.

26-1 capital letters are still available, plus 26 lowercase letters, which equals 51.

[SM-Count]

Wouldn't it be 26*(50...)and so on? Since you have to remove both the capital letter & the lower case letter from the first pick.

26-1 capital letters are still available, plus 26 lowercase letters, which equals 51.

'A' & 'a' are the same letter.