Show how the probability that an 8 character password contains exactly 1 OR 2 integers is .630 A password is 8 characters long. Each character can contain 26 lower case or 26 uppercase letters or a integer from 0-9. What is the probability that an 8 character password contains exactly 1 OR 2 integers? 
Please explain as the book answer is P = .630 
Assuming that I could use (52^6*10^2+52^7*10)/62^8 was incorrect. This is the sum of the product between the possibilities of having one integer in the password and having two integers in the password; 62^8 is the total possibilities given each character can be a lower/uppercase or integer.
 A: Hint: The number of passwords with exactly $2$ digits is $\binom{8}{2}(10^2)(52^6)$.
This is because the locations of the $2$ digits can be chosen in $\binom{8}{2}$ ways. For each of these ways, the $2$ chosen locations can be filled with digits in $10^2$ ways. And for each of these ways, the remaining $6$ slots can be filled with letters in $52^6$ ways. 
Your calculation did not take into account the fact that there are quite a few choices for the location of the two digits. 
A: First of all, the number of passwords we can form is $(26+26+10)^8=62^8$
Now how many of these passwords have $1$ number in them? The seven letters can be $52^7$ different combinations, and the number can be $10^1$ different numbers, giving a total of $52^7*10$ different combinations. We also have to remember that the password can be in $1$ of $8$ spots, so we need to multiply that number by $8$, so we get $52^7*10*8=80*52^7$
Now how many passwords have $2$ numbers in them? Using the same logic as above, we can get $52^6*10^2*\frac{8!}{2!6!}$.
The total number of passwords is now $80*52^7+52^6*10^2*\frac{8!}{2!6!}$, making our total probability  $$\frac{80*52^7+52^6*10^2*\frac{8!}{2!6!}}{62^8}$$
