8 character password Everyone is asked to create a new 8 character password with at least one number and exactly one special character with the remaining characters being lowercase letters. How many possible passwords are available?
I don't get how to find with restrictions. 
26 letters 
10 numbers 
9 special characters. 
Please help me. 
 A: We first count the number of 8-character passwords with no numbers at all, but exactly one special character. We


*

*Choose the special character: $9$ ways

*Place it: $8$ ways

*Choose a letter for each of the remaining positions: $26^7$ ways


So we can do it in $9 \cdot 8 \cdot 26^7$ ways. Now, we count the number of 8-character passwords with exactly one special character and no restrictions on the number of numbers. We


*

*Choose the special character: $9$ ways

*Place it: $8$ ways

*Choose a letter or a number for each of the remaining positions: $(26 + 10)^7$ ways


The number of such passwords is thus $9 \cdot 8 \cdot (26+10)^7$. If we want to count the number of passwords with at least one number, we subtract the number of passwords with no number from the total number of passwords. Hence the answer is
$$9 \cdot 8 \cdot (26+10)^7 - 9 \cdot 8 \cdot 26^7 = 5063929482240$$
A: I am not sure whether this is the best or easiest way but here is one way to attack the problem.  I will use L to represent an arbitrary letter, 9 for an arbitrary digit, and * for a special character.  
Case 1 – Exactly one number
Case 1a – The letter is in position 7 and the special character is in 8 i.e. LLLLLL9* so, there are 26^6 x 10 x 9 possibilities.  
Back to the general case 1, how many other arrangements are there?  The digit can be in any of 8 positions and the special character in any of the other 7.  So 8 x 7 x the number of 1a cases.
Case 2 – Exactly two numbers
Case 2a – the pattern LLLLL99* has 26 ^ 5 x 10 ^ 2 x 9 possibilities.  
Back to general case 2, how many other arrangements?  The two digits can be in 8 x 7 / 2 places and the special character in one of the remaining 6.  
Case 3 etc.  Continue in the same way.  You will need some simple combinatorics e.g. how many arrangements of three digits, four digits, etc.  
The rule appears to allow case 7: 0 letters, 7 digits and 1 special character so don’t forget that.  
