how many choices are there?

26 letters and 10 numbers gives you 36 possible characters and if you are only sticking with the 6 character limit and no more I think (36^6 - 36^5) seems likely. (give or take a few million just to be safe.)

I came up with 2,116,316,160 possible 6 character combinations but I could be wrong though.

Little off the subject but the same principals.

I had a young engineer that was working with a salesman (suposable a simple devert conveyor)

This conveyor ballon out of shape after just a week.

The scope this conveyor was to devert into 7 lanes with any combinations open/closed. Nice and discreet right.

Well they were going to use devert paddles staged.

When I found out about it, I first talked to the EE dept. and the young engineer that had said it would be no more than 16 combinations at the most of possiblities? off the top I said possiblities is more like 7^7. They budged and said well maybe 36 at the most.

I shook my head and offered up a more realistic solution, which as they dug into it realized they folly.

I often thought of the posibilies of that. Without doing a truth table. Where did you come up with 36^6 - 36^5 of posibilities, I don't see it, But with those grounds the possibilties in my case would be 7^7 - 7^6 = (705894) This would be a good puzzle question posed.

p911

There are 2 solutions depending on possible repetition or not:

1- without repetition the number of possible combinations for "k" signs of "n" is N= n*(n-1)*(n-2)*....*(n-k)

2- when same sign can appear several times in a combinations then the result is different N1= (n+k-1)*(n+k)*(n+k-1)*....*(n-1)

In your case with n= 36 and k=6 N= 1,402,410,240

N1= 3,237,399,360

If the order has not to be repeated then the above values have to be divided by 6!=720.

typical usernames are not case sensitive

there can't be negative values

I got as far as 36⁶, but couldn't see the reason to subtract 36⁵. If the password is case sensitive, it would be 62⁶, or less if there are various other restrictions such as at least one letter and one digit.

On the diverter conveyor, shouldn't it be 2⁷ = 128, or 127 if the null set is disallowed?

No null set is allowed

My reasonings where that since numbers do not have upper and lower case or capitalization I just went with the 26 base letters and left out the possible sub variations. The OP did not specify so I took the lowest number of characters which was 36.

Subtracting the 36^5 is necessary since all combinations of characters that are less than 6 total don't count.

But yes over all this question can be argued a number of ways and each possible assumed or implied method will come up with a different total. Odds are this thread could get drawn out pretty far because of the open ended way to interpret the base question.

Normal rules for a secure user name

At least contain 1 upper case

At least contain 1 lower case

At least contain a number

so that will make it

26 * 26 * 10 * 62^3 = (for upper or lower case )

1 611 097 280 (if I can trust my cheap Chinese calculator )

I just logged out then came back and logged in using my screen name log in but with different upper and lower case letters than normal. So it would seem that on this site like many others the upper and lower case letter values have no different meaning so regardless of capitalization or not it would appear that this site just uses the 26 base letters only.

So for the argument here we are stuck with the 26 base letters and 10 numbers that the log in system recognizes. Upper and lower case are invalid, "a" and "A" are seen as the same character, and treated equally by this system which then only gives us 36 base characters it works with not 62.

Ah, but you're forgetting there's one codebreaking method that often yeilds the solution to virtually any encryption technique within only a few minutes.

http://xkcd.com/538/

I'm not trying to do a hack

just wondering how many combinations there are & how to figure the answer out

I'm not sure any of the answers are right so far. For instance, my initial 36⁶ does not weed out duplicates that occur from repetition. I'm not sure yet, but I don't think subtracting 36⁵ accounts for this, either. It would be a good challenge question, but likely to attract ill-informed but unclear and contentious answers. Also not sure that 26² x 10 x 62³ (or 26² x 10 x 36³) accounts for the ways the specified l.c., u.c., and digit might be placed in the username. Trickier than first meets the eye....

it already Looks like a challenge question to me

what would the nature of the duplicates be? there are no null places.

You also need to specify a maximum amount of characters, and here is a calculator:

http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

and if all else fails, WIKIPEDIA!!!!

http://en.wikipedia.org/wiki/Combinations

http://en.wikipedia.org/wiki/Permutation

If no characters may appear more than once (upper or lower case)

the number would be

36! / 30! / 6! = 36 * 35 * 34 * 33 * 32 * 31 / 5 /4 / 3 /2 = 3 344 528

Sorry--no cigar. (Even left out the 6 from the 6! part.)

Oops

There are several interpretations as follows:

1- There are only 36 symbols available and the question is how many different groups of 6 can be build. The answer depends of the possibility to ask for possibilities which have same symbols in different orders (a) or not (b).

2- The 36 symbols can be used for every position in the group of 6 (c) which means that the bank contains 6x36 = 216 symbols.

a- Let us look at each place in the 6

1^st place 36 symbols are available

2^nd place only 35 remain since 1 was used for the 1^st place

And so on till the last place for which only 36-(6-1) symbols are still available.

The result is thus 36*35*34*33*32*31 =

b- All the groups of 6 having same symbols are obtained by permutations. If we want to avoid it we have to divide above result with the possible number of permutations which is 6! (factorial) = 1*2*3*4*5*6 = 720. The result will be N' = 1947792

c- Let us proceed as above the difference is that for each place 36 symbols are available so that the table becomes:

If we again want to avoid groups with same symbols but in different orders we have to divide as above by 720 and obtain the result as N1' = 3023308.

I hope now to have explained the logics behind the values.

Nick

Thanks Nick that helped ...

My guess is that the last is probably the most correct

I was trying to come up with a Username for Gmail & began to wonder about the number of possibilities...

There are other rules, but none of these are explicit