Here's the question:
How many six-character passwords can be built with lowercase letters and numbers, given that at least one of its characters is a number?
Here's my answer:
$$10\dbinom{6}{1}36^5$$
- $10$ for the number possible choices for digits
- $\binom{6}{1}$ where one location in the $6$-character string is being chosen for the digit to be placed
- $36^5$ for remaining character combinations ($26$ letters + $10$ digits)
Here's the actual answer:
$$36^6 - 26^6$$
- $36^6$ for all strings of length $6$
- minus $26^6$ for the number of strictly alphabetical passwords
I get why the solution works, but I can't see where my solution went wrong; it's quite a bit bigger than the answer. Can someone explain why my answer is wrong?