Probability of Picking 3 Correct Digits on a 10,000 Combination Lock I have a combination lock question that's different from the other ones I've searched for here.  Here goes:
There's a combination lock that goes from 0000 to 9999. The correct combination is 1234.
What is the probability of guessing exactly 3 of the combination's actual digits (1,2,3,4), without regard for the guessed digits' order?  (For example, 341, 243, 412, 241, etc. would be acceptable 3-digit guesses).
My question is I'm not sure which approach to take. What I have so far (and I'm not even sure if it's right) is:

*

*For the first digit guess (which can range from 0 to 9), 4 digits are allowed (1,2,3 or 4), so: 4/10

*For the second digit, there are 3 choices are left (because we chose one from the set of 1,2,3,4 already), so 3/10

*For the third digit, there are 2 choices left (because we chose two from the set of 1,2,3,4 already), so 2/10.

*So, 4/10 x 3/10 x 2/10 = 3/125

Is this in the ballpark?  Thank you for any hints!
 A: Revised answer according to considered  interpretation of the question
I have interpreted the question as " Suppose the lock has a $4$ digit code with $4$ distinct digits, say $1234$, unknown to me, what is the probability that I get exactly $3$ of the $4$ digits right, even though the digits might be in wrong positions, eg for $1234, 2138, 0432$
etc would be considered right.
Of the $10$ digits, $4$ are used, $6$ are not, so in order to get exactly $3$ digits right,
$Pr = \Large\frac{\binom43\binom61}{\binom{10}4} = \frac 4{35}$
A: You can work step by step. First digit can be in 1,2,3,4 (P=0.4) or not (P=0.6)
Second digit can be in 1,2,3,4 and different from first one, or not.
After 2 digits, number of distincts digits in 1,2,3,4 can be 2 (P=0.12) or 1 (P=0.52) or 0 (P=0.36)
If number of distinct digits is 2, the 3rd digit can be in 1,2,3,4 , and different to first 2 ones (P=0.2), or it can not change the count of distinct digits (P=0.8)
Etc, etc
Final result : With 4 digits, you may have 4 distinct digits all in 1,2,3,4 : P=0.0024,
or 3 distincts digits in 1,2,3,4 (P=0.072)
or 2 distincts digits in 1,2,3,4 (P=0.354)
or 1 distinct digit in 1,2,3,4 (P=0.442)
or 0 distinct digit in 1,2,3,4 (P=0.1296)
