Five digit numbers are formed with 0,1,2,3,4. find the probablity of getting 2 in ten's place and 0 in the units place always

I assume that no digit is used more than once...

In the ten thousand's place, you can have 1, 2, 3 or 4. (4 possibilities)
In the thousand's place, you can have 0, 1, 2, 3 or 4. (now, you used one of the 5 above, this becomes 4 possibilities)
In the hundred's place, you can have 0, 1, 2, 3 or 4. (you used two above, which makes 3 left)
In the ten's place, you can have 0, 1, 2, 3 or 4. (2 left)
In the unit's place, you can have 0, 1, 2, 3 or 4. (1 left)

In all, you have 4x4x3x2x1 = 96 possibilities.

Now, how many of these contain 2 in the ten's place?
_ _ _ 2 _

Applying the same as above, we get: 3x3x2x1 = 18 numbers.

So, P(2 is in the ten's digit) = 18/96

Can you do the other one now?

And then, P(2 is in the ten's digit place and 0 in the units place) = P(2 is in the ten's digit) x P(0 is in the units place)