ah... I see what you're saying...
On Feb 9, 8:56 pm, Dave wrote:
> @SVIX: According to my calculation, this gives
> 2,992,430,052,218,880,000, almost 12 times the correct answer,
> 251,471,033,958,144,000, that I gave earlier in
> postinghttp://groups.google.com/group/algogeeks/msg/bb2269736a997
@SVIX: According to my calculation, this gives
2,992,430,052,218,880,000, almost 12 times the correct answer,
251,471,033,958,144,000, that I gave earlier in posting
http://groups.google.com/group/algogeeks/msg/bb2269736a997419. This is
because you are counting some passwords multiple times. Consid
1. there should be atleast 3 capital letters
2. atleast 3 small letters
-> 6 spaces gone for these, with repetitions allowed.
for 3 spaces, we have 26^3 possibilities, and they can be arranged in
10C3 ways...
for the next 3, they can be arranged in 7C3 ways
3. atleast 2 numbers 0-9
now, 4 spac
No. That is too large, at 2,087,438,895,360,000,000. Analyzing your
expression, 26C3 is the number of ways to choose 3 different letters,
but the letters can be the same. The number of combinations of 3
letters with repetitions is 26^3. So that aspect of your formula is
too small. However, you can'
The following combinations of capital letters "C", lower case letters
"l", and digits "d" are possible:
C l d Number possible
3 3 4 10C3 * 26^3 * 7C3 * 26^3 * 4C4 * 10^4
3 4 3 10C3 * 26^3 * 7C4 * 26^4 * 3C3 * 10^3
3 5 2 10C3 * 26^3 * 7C5 * 26^5 * 2C2 * 10^2
4 3 3 10C4 * 26^4 * 6C3 * 26^3 * 3C
