Problem Solving with Combinations question?

My guess is you are being thrown by the English rather than the mathematics. So for a) given that the password has 8 characters, a password with at least two letters means that of all the possible  (26 +10)^8 passwords, we have to remove the passwords with less than two letters. So the task then becomes finding the number of 8 character passwords with less than two letters, which means one letter and all the rest are numbers. So the single letter has 26 possible values in 8 possible positions. And every remaining character must be a number with 10 possible values. 

So you have 26+10=36 characters to choose from.

a) Here, best to evaluate the opposite: Out of ALL the ways, how many

have no letters and at most 1 letter?  [do you see how this relates to your question?]

1) ALL the ways: 36 * 36 * ... * 36  [8 times]. Gives you?

2) No letters: 10*10*...*10  [8 times] = ?

3) Only 1 letter: 26 choices which can go at 10 places, then  the remaining seven places have numbers: (26C1 * 10C1)*10*10*...*10 [7 times] = ?

To finally answer your question: (1) - (2) - (3) = ?

So, what answer does this finally give? Answer that and we will see if you did it okay. Then we'll proceed to the other questions.

Total number of passwords is 36^8.

Number of passwords containing no letters is 10^8.

Number of passwords containing exactly 1 letter is

10^7*10*26 = 26*10^8.

The number of passwords containing at least two letters is

36^8 - 27*10^8 

= 2.8211 x 10^12 - 0.0027 x 10^12 = 2.8184 x 10^12.

That's my answer for (a).  The strategy for (b) is quite similar.

For (c) you could just take the two minuends from (a) and (b) but subtract BOTH from 36^8.