**Password Calculations: mathematic permutation and combinatorics.**

Password permutations allow you to calculate exactly how strong any password is, based upon what the password contains.

The variables that should be addressed in calculating password strength * are the length of password, and the characters it contains.* The length of the password is the first variable in the calculation. The second variable is the composition, or contents, of the password.

**Example 1:** For example, let’s say I told you to create a password with the length of TWO characters, and to be comprised of only UPPERCASE letters.

The resulting calculation would be the possible number of characters that can be utilized (26 uppercase letters in the alphabet) raised to the power of the password length; in this case, the length is 2 digits, so you raise 26 to the quantity of 2.

**Let n = possible number of characters**

**Let ^b = length of password**

**Therefore, password permutation and combinatorics = n^b**

In this case, n=26 & b = 2, resulting in 26^2.

The resulting calculation is therefore, 26 multiplied by 26. (26*26)

To summarize that’s the possible quantity of characters that can be used (26 possible characters) risen to the power of the length of the password (TWO digits). Since in this case, the quantity of characters that can be used is TWENTY SIX, and the length of the password is TWO, you raise 26 to the power of 2. 26^2 = multiply 26 by 26.

So, the length of the password dictates how many times you multiply the possible quantity of characters by itself.

**Example 2:** Now let’s say the password length is three, comprised of lowercase letters. Since both lowercase and uppercase each have 26 possible characters to input, The calculation would appear as follows: 26*26*26. Resulting in 17,576. That means, that your three digit password, comprised of ONLY lowercase letters, would have 17,576 possibilities.

**Example 3:** Let’s say you have a password length of 3, but you can only use NUMBERS. So, the resulting mathematic combinatorics would be: 10*10*10 = 1,000.

**Example 4:** Imagine your system had a requisite demand for a password with 4 numbers, but you could ONLY utilize each number ONCE. This changes the formula, because the quantity of characters you can use diminishes each time.

The resulting calculation in this case would be: 10*9*8*7 = 5,040

**Imagine the possibilities of password combinatorics and mathematic permutations.**

By making wise utilization of the theories discussed, you can calculate the mathematic permutation and combinatorics for ANY password mechanism that can be composed.

Consider this, how many possibilities would there be for a password with the length of 10 digits comprised of uppercase, lowercase, and any number? Consider your original formula that password permutation and combinatorics = n^b. In this case, n=62 (possible number of characters: 26 lowercase letters + 26 uppercase letters + 10 numbers) and b = 10 (10 digits).

Therefore the resulting calculation would be 62^10.

Now, imagine, how many possibilities would there be for a 40 digit password, that can utilize lowercase, uppercase, numbers, and up to 15 special characters? What about a 150 digit password?