License Plate Probability: No Repetition Problem: Find probability of a plate having an A and a 1, no repetition allowed.
I haven't done these types of problems in a while, and originally did it assuming repetition was allowed before I noticed the fine print saying it isn't.
Plates are 3 letters followed by 4 numbers. 
Just to show my thought process, originally when I assumed repetition was allowed, I had $(1-(25/26)^3)*(1-(9/10)^4)$ for getting a plate with at least an $A$ and a $1$ (1 - prob no A's AND no 1's).
For no repetition, I ended up with $(3/26)*(4/10) = 3/65$. If you get an A, the other 2 letters don't matter and there are 3 orderings. If you get a 1, the other 3 numbers don't matter and there are 4 orderings. but I'm not confident that's correct. Confirmation of that and/or help would be appreciated.
Edit: I will update and select an answer after I get some clarification on the problem in a few hours. Thank you both, you were extremely helpful!
 A: It seems like you got the correct answer but your explanation sounds flawed - after selecting one A, you cannot "not care" about the other two letters - you have to guarantee that they are different from A and each other.
For exactly one A, you simply ask yourself: How many possible valid license plates can I construct? There are three possible schemes for one A:
AXY XAY and XYA

where X denotes a non-A letter and Y is a non-A non-X letter. You cannot "not care" about the X - it has to be a non-A letter since otherwise it could possibly violate the "exactly one A" rule.
The A is fixed, you can only choose it in one way. The X can be selected in 25 different ways, the Y in 24 (neither A nor X) so you have $1 \times 25 \times 24 = 600$ possible ways of choosing three letters of which two are non-A and one is A. Now multiply this number by the total number of distinct orderings, $3$. You get $3 \times 600 = 1800$, which is the number of possible arrangements for all valid letter choices. Dividing this by the total number of orderings, $26 \times 25 \times 24$, gives us the probability of a random ordering conforming to the one A, two non-As rule. So the final probability is:
$$\frac{3 \times 25 \times 24}{26 \times 25 \times 24} = \frac{3}{26} \quad \text{or about} \quad 11.54\%$$
For the other case, just try the same strategy as above. (How many valid choices of numbers with exactly one 1 are there? How many ways can I arrange them? What is the total number of arrangements?) The final probability for it having exactly one A and exactly one 1 is the product of both probabilities.
A: There are $26^3 \cdot 10^4$ different plates, so:
$$|\Omega| = 26^3 \cdot 10^4$$
or, if we assume that there are no repetitions within plates:
$$|\Omega_1| = 26\cdot 25\cdot 24 \cdot 10\cdot 9\cdot 8\cdot 7$$
Now we want plates without repetitions, but with A and 1, so first we need to select two dofferent letters different than A (the order matters) and a place where we put A. Then we select 3 different numbers different than 1 and select a place, where we put 1. 
$$|A| = 25\cdot 24 \cdot {3\choose 1} \cdot 9\cdot 8 \cdot 7 \cdot {4\choose 1}$$
Finally the probability obtaining the license plate with A and 1 and without repetitions is :
$$P(A)=\frac{|A|}{|\Omega|}=\frac{25\cdot 24 \cdot 3 \cdot 9\cdot 8 \cdot 7 \cdot 4}{26^3 \cdot 10^4}$$
or, for plates with no repetition,
$$P(A)=\frac{|A|}{|\Omega_1|}=\frac{25\cdot 24 \cdot 3 \cdot 9\cdot 8 \cdot 7 \cdot 4}{26\cdot 25\cdot 24 \cdot 10\cdot 9\cdot 8\cdot 7} = \frac{3}{65}$$
