The password for a computer account has to consist of exactly eight

Hi pkk,

In permutations and combinations, if repetition is allowed, then each event is independent from one another. For example, if we are picking a 4 digit code from the numbers 0-9 WITH repetition, then for each selection we can pick any number for each digit. The choices made for the first digit will not affect the choice for the second, third and fourth digits. In this case, the number of permutations would be \(10^4\). If we're dealing with combinations (where the order doesn't matter) (i.e. the code 1123 is equivalent to the code 3112, etc.), then the number of combinations is \(\frac{(r+n-1)!}{(r!)(n-1)!}\), where r is the number of digits in our code, and n is the amount of different numbers we can choose (0-9). So in this example the number of combinations with replacement is: \(\frac{(4+10-1)}{(4!)(10-1)!}=\frac{13!}{(4!)(9!)}\).

If repetition is not allowed, then each choice made restricts the subsequent choices. In our example, if I were to choose a code of 4 digits without repeating any digits, then the total permutations would be 10!/6! = 10*9*8*7. In other words, I can choose any number for the first digit, any of the 9 remaining numbers for the second digit, any of the 8 remaining numbers for the third digit, and any of the 7 remaining numbers for the last digit.

If we're talking about combinations (order is not important), then we just have to take into account all the ways a group of 4 numbers can be arranged, which is 4!. So the number of combinations without replacement would be \(\frac{10!}{(6!)(4!)}\).

For a nice detailed explanation with more examples, check out
https://www.mathsisfun.com/combinatorics ... tions.html


Cheers,