Re: stat password problem.

Quote:

Originally Posted by

**rcmango** there are 26 possible letters in the english alphabet, pick passwords that don't contain any repeated letters. passwords are 5 letters each, how many passwords are to choose from?

I'm not sure how to solve this problem.

However, I know the answer is: 7,893,600

what type of stat problem is this, and what are the keywords to identify it, also the equation to solve it. Thankyou.

Combinatorial

CB

Re: stat password problem.

Is there an equation to help me with this problem?

Re: stat password problem.

Quote:

Originally Posted by

**rcmango** Is there an equation to help me with this problem?

By repeats do you mean that no letter appears twice, or do you mean no two consecutive letters are the same?

If the first: The first letter may be any of the 26, for each case the second may be any of the remaining 25, ... Multiply these together to get the number of passwords.

If the second: The first letter may be any of the 26, the second any of the remaining 25, the third any of the 25 not the same as the second, ... multiply these together to get the number of passwords.

CB

Re: stat password problem.

Let suppose that n is the size of alphabet. The number N of k letters passwords...

a) without any sort of limitation...

(1)

b) without repetead consecutive letters...

(2)

c) without repetead letters at all...

(3)

If n=26, k=5 we have...

a) N= 11,881,376

b) N= 10,156,250

c) N= 7,893,600

These results merit some comments. It is evident the loss of security passing from a) to b) and even more from a) to c). One can ask: what is the reason of that?...

The probable answer may be: the reason is that if someone wants to 'catch' a 'random word' he tends to exclude words with any sort of repetion, so that a 'clever idea' is to adopt passwords with repetead letters... Very clever idea indeed!... but what does it happen if a 'not very clever hacker' realizes this 'clever idea' has been adopted?... in this case, supposing the 'strategy' b), the number of possible passwords is reduced to...

N= 11,881,376 - 10,156,250 = 1,725,126

A very clever idea indeed!(Headbang)...

Kind regards