If I create a $10$ digit password with the following requirements:
- At least one uppercase letter A-Z : $26$
- At least one lowercase letter a-z : $26$
- At least one digit from $0-9$ : $10$
- At least one common symbol $(\#,\$,\%,$ etc) : $32$
By inclusion-exclusion, I can calculate I have ~ $3.2333\mathtt E+19$ possible combinations
However, if I change one of the requirements to at least TWO digits $0-9$, how can I calculate the possible combinations?
Best Answer
You have to choose 10 letters, and 2 of them must be digits. Furthermore, there must be one each of a lowercase letter, an upper case letter, and a common symbol. For the others, there are 5 choices to be made, and these are to be made from $26+26+10+32 = 94$ characters.
This gives
$94^5 \approx 7.339e9$
choices for the the 6 other characters that do not have to be digits. And for the digits, there are 2 choices from 10 characters. So this gives
$ 10^2 = 100$
choices. And for the one each of lower-case letters, upper-case letters, and common symbols there are
$ 26 * 26 * 32 = 21632$
choices.
Now lastly, there are $10!$ permutations of these 10 characters, so the total number of combinations of these characters is:
$26*26*32*100*94^5 * 10! \approx 5.76e22$