Math Problem

Dystopia · **Joined:** Jan 2007 **Posts:** 2317 **Location:**

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 · **Posted:** Sun Feb 17, 2008 7:52 pm

dont ask me^^ i suck at math

bLuE_fLaMe · **Joined:** Apr 2007 **Posts:** 629 **Location:**

ask on yahoo answers, some nerd will know it

SM-Count · **Joined:** Jan 2008 **Posts:** 2761 **Location:** /wave

Stress · **Posted:** Sun Feb 17, 2008 7:58 pm

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 · **Joined:** Jan 2007 **Posts:** 2317 **Location:**

Stress wrote:

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 · **Joined:** Apr 2006 **Posts:** 3216 **Location:** wat

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

Azilius · **Joined:** Oct 2006 **Posts:** 4236 **Location:** CS:GO

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 · **Joined:** Jan 2007 **Posts:** 2317 **Location:**

Azilius wrote:

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 · **Joined:** Oct 2006 **Posts:** 4236 **Location:** CS:GO

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 · **Joined:** Jan 2008 **Posts:** 2761 **Location:** /wave

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 · **Joined:** May 2007 **Posts:** 4441 **Location:** Life

SM-Count wrote:

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 · **Posted:** Sun Feb 17, 2008 8:44 pm

SM-Count wrote:

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 · **Joined:** Aug 2007 **Posts:** 2587 **Location:** OTL&OG

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 :oops:

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 :banghead:

Good luck on solving it lol^^

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

NuclearSilo · **Posted:** Sun Feb 17, 2008 9:18 pm

26^9

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

Inuyasha584 · **Joined:** Jun 2007 **Posts:** 1491 **Location:**

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

NuclearSilo · **Posted:** Sun Feb 17, 2008 9:38 pm

^ = power

Inuyasha584 · **Joined:** Jun 2007 **Posts:** 1491 **Location:**

salmissra · **Joined:** Nov 2007 **Posts:** 1487 **Location:**

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 · **Joined:** Sep 2007 **Posts:** 155 **Location:**

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 · **Posted:** Mon Feb 18, 2008 11:20 am

No reply wrote:

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 :shock:

Simplifique · **Joined:** Sep 2007 **Posts:** 372 **Location:**

No reply wrote:

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 · **Joined:** Sep 2007 **Posts:** 155 **Location:**

Quote:

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 · **Posted:** Mon Feb 18, 2008 12:22 pm

explain the formula -_-

No reply · **Joined:** Sep 2007 **Posts:** 155 **Location:**

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 · **Posted:** Mon Feb 18, 2008 1:14 pm

Yup. I double checked. Shadowman is right

SM-Count · **Joined:** Jan 2008 **Posts:** 2761 **Location:** /wave

No reply wrote:

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 · **Joined:** Sep 2007 **Posts:** 372 **Location:**

SM-Count wrote:

No reply wrote:

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 · **Joined:** Sep 2007 **Posts:** 155 **Location:**

Quote:

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 · **Joined:** Jan 2008 **Posts:** 2761 **Location:** /wave

No reply wrote:

Quote:

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.

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