Password Permutation/Combination Problem So here's the problem:
At a certain company, passwords must be from 3-5 symbols long and composed of the 26 letters of the alphabet, the ten digits 0-9, and the 14 symbols !,@,#,$,%,ˆ,&,*,(,),-,+,{, and }.
a. How many passwords are possible if repetition of symbols is allowed? b. How many passwords contain no repeated symbols? c. How many passwords have at least one repeated symbols?
I got 50^5 (312,500,000) for A and 50! for B, and I know the formula to figure out C is 
|at least one|=|all|−|none|
but if I do A - B, I end up with a negative number, which I know isn't correct. What am I doing wrong?

For a, you are assuming only 5 character passwords. You need to add in the 3 and 4 character cases. In this case, a would be $50^5+50^4+50^3$, yes.
For b, you are considering only 50 character passwords which are way too long. 50 choose 5 would give you the 5 character case that is much smaller as it would be $\frac{50!}{45!}$ to compute this. You also have to add the cases for 3 and 4 character cases here. (I forgot that order would matter here.)
Let's consider a 3 letter password without repetition for a moment. How many choices are there for the first character? The second character the third character? This would be $50*49*48$ whereas you going for $50!$ would be presuming the password to use all the characters which isn't right. This is how we'd get the $\frac{50!}{47!}$ here for the 3 character case.
For c, you could take the value for a minus b once they are correct.
