Number of 6-character passwords containing at least one number – why is the answer wrong

Question

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?

Answer 1

For example, the string 11AAAA is counted twice in your solution, once when your "minimum guaranteed digit" is the first one, and once -- when it is the second one...