# Combinatorics

• June 16th 2009, 02:34 PM
Keep
Combinatorics
This is the question. A computer has a password composed of 6 characters. The first 2 letters must be small letters, the rest can be either numbers or small letters.

a) How many different passwords are possible?
b) How many passwords are there that have no repeated characters?
• June 16th 2009, 02:44 PM
Plato
Quote:

Originally Posted by Keep
This is the question. A computer has a password composed of 6 characters. The first 2 letters must be small letters, the rest can be either numbers or small letters.

a) How many different passwords are possible?
b) How many passwords are there that have no repeated characters?

At least, tell us how far you have gotten.
• June 16th 2009, 02:58 PM
Keep
Quote:

Originally Posted by Plato
At least, tell us how far you have gotten.

First, the computer has a password of 6 characters. That means there is a possibility of up to 6! = 720 right? If no character is repeated, then we will still have 720 ways of writing the passowrd?
• June 16th 2009, 03:11 PM
Plato
Quote:

Originally Posted by Keep
First, the computer has a password of 6 characters. That means there is a possibility of up to 6! = 720 right?

No. Because characters can be repeated. Only the first two must be alphabet.
So the are $26^2$ possiblities for the first two.
Then the next four can be any one of $26+10=36$.
So what is the answer to part a)?
• June 16th 2009, 03:24 PM
Keep
So a is 36 + 676. Just to clarify on how you got 26. It is because of the number of ways in which the numbers can be entered which is 4! plus 2 for the characters or? Why did you raise it to the power of 2 then? And lastly, why did you add 10 to 26?
• June 16th 2009, 03:39 PM
Plato
Quote:

Originally Posted by Keep
So a is 36 + 676. Just to clarify on how you got 26. It is because of the number of ways in which the numbers can be entered which is 4!

Absolutely not!
Surely you know that there are 26 characters in the English alphabet.
Do you have the basic ‘common body of knowledge’ to do this problem?
For example: How many digits are there?
• June 16th 2009, 03:59 PM
Keep
Quote:

Originally Posted by Plato
Absolutely not!
Surely you know that there are 26 characters in the English alphabet.
Do you have the basic ‘common body of knowledge’ to do this problem?
For example: How many digits are there?

Thank you for your desire to help. Unfortunately, I haven't understood it very well because the resources with which we were provided was extremely limited. Nevertheless, I will try to understand it. First, the 26 is the number of characters in the English raised to 2 which is the first two characters which are alphabets. Then you add 26 to 10. 10 is the number of digits. This gives you 36 and you raise this to the power of 4. Then you add the $26^2$. In other words you do $26^2 + 36^4$
• June 16th 2009, 04:09 PM
Plato
Quote:

Originally Posted by Keep
. In other words you do $26^2 + 36^4$

Actually there are $(26)^2(36)^4$ as an answer for part (a).
You explain that!

Part (b), there are no repeats.
The answer is $(26)(25)\left(^{34}\mathcal{P}_4\right)$. WHY?
• June 16th 2009, 04:29 PM
Keep
Quote:

Originally Posted by Plato
Actually there are $(26)^2(36)^4$ as an answer for part (a).
You explain that!

This is because there are $26^2$ ways for the first 2 characters and $36^4$ for the remaining 4 characters which can either be digits or numbers (26+10)?

Quote:

Originally Posted by Plato
Part (b), there are no repeats.
The answer is $(26)(25)\left(^{34}\mathcal{P}_4\right)$. WHY?

This is because there are in total 26 characters and there are 26 ways for the first letter and 25 ways for the second letter then 34 ways for the remaining 4 characters: 10 digits and 24 letters?

Any more explanation is appreciated, and thank you.
• June 16th 2009, 05:09 PM
Plato
Quote:

Originally Posted by Keep
Any more explanation is appreciated, and thank you.

Think about it. The first two are alpha.
The next four are either alpha or digit.
So the last four can be chosen in (34)(33)(32)(31) ways.
• June 16th 2009, 05:20 PM
Keep
Ok, I understood it now. Thank you once again for your patience and help! Understanding is always better than getting just an answer.
• June 16th 2009, 11:02 PM
Swlabr
Quote:

Originally Posted by Plato
$\left(^{34}\mathcal{P}_4\right)$.

Hi,

I'm just wondering what that means - I don't think I've come across that notation before.

Is it just $\left(^{n}\mathcal{P}_i\right) = n(n-1) \ldots (n-i+1)$?

Thanks.