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?
 A: Number of passwords with (possible) symbol-repetition:
$(50^3)+(50^4)+(50^5)=318875000$

Number of passwords without any symbol-repetition:
$(50\cdot49\cdot48)+(50\cdot49\cdot48\cdot47)+(50\cdot49\cdot48\cdot47\cdot46)=259896000$

Number of passwords with at least one symbol-repetition:
$318875000-259896000=58979000$
A: 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.
A: It's also possible to think of "nothing" as a sort of symbol of its own, one that has more restrictions on where it can be put, and look at it in another way. This ends up being exactly the same as considering each password length separately.
With or without symbol repetition:
$$
\begin{align}
N &=
\underbrace{50^3}_{\text{1st-3rd}} \cdot (\underbrace{1}_{\text{Two blanks, or,}} + \underbrace{50}_{\text{(4th filled and}} \cdot \underbrace{51}_{\text{5th filled or blank)}} ) \\
&= 50^3\cdot(1+50\cdot(1+50)) \\
&= \underbrace{50^3}_{\text{3 digit}} + \underbrace{50^4}_{\text{4 digit}} + \underbrace{50^5}_{\text{5 digit}} \\
&= 318,875,000
\end{align}
$$
Without symbol repetition:
$$
\begin{align}
N_\text{no rep} &= 
(\underbrace{50\cdot 49 \cdot 48}_{\text{1st-3rd}}) \cdot (\underbrace{1}_{\text{Two blanks, or,}} + \underbrace{47}_{\text{(4th filled and}} \cdot \underbrace{47}_{\text{5th filled or blank)}} ) \\
&= (50\cdot 49\cdot 48)\cdot(1+47\cdot(1+46)) \\
&= (\underbrace{50\cdot 49 \cdot 48}_{\text{3 digit}}) + 
   (\underbrace{50\cdot 49 \cdot 48 \cdot 47}_{\text{4 digit}}) + 
   (\underbrace{50\cdot 49 \cdot 48 \cdot 47\cdot 46}_{\text{5 digit}}) \\
&= 259,896,000
\end{align}
$$
With at least 1 symbol repeated:
$$
N_\text{rep} = N - N_\text{no rep} = 58,979,000
$$
