# How can I calculate the number of potential combinations in a password?

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 digits 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? - By inclusion-exclusion again. There are just more terms. – Qiaochu Yuan Aug 27 '10 at 4:04 You can take your previous answer, compute the number of passwords that had exactly one digit, and subtract it. – Arturo Magidin Aug 27 '10 at 16:09 So since there are 3.2333E+19 possible combinations remaining, and each of those has at least 1 digit, and the most digits it can have is 7, wouldn't the answer be 6/7's of the 3.2333E+19 = 2.77143E+19? – user1524 Aug 27 '10 at 19:11 ## 1 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\$

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I'm confused by this method. I started with 94 possible characters with no restrictions as 94^10 = 5.38615E+19 total combinations possible, and then subtracted the restricted sets. This number is higher than my start point for some reason. – user1524 Aug 27 '10 at 19:14