Question 178085
you should be able to get 4!/2! unique numbers out of these digits.
if all the digits were different, the formula would be 4!
since 2 of the digits are the same, you need to divide by 2!
your answer should be:
4!/2! = 4*3*2*1/2*1 = 24/2 = 12
if 3 of the digits were the same, your answer would have been 4!/3!
that would equate to 4*3!/3! = 4
in your original case, the answer is 12.
here's how they would break down.
1399
1939
1993
3199
3919
3991
9139
9193
9319
9391
9913
9931
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fyi, if you changed the 3 to a 9, then you would have had 3 digits the same and the answer would  have been 4!/3! = 4
the selections would then have been:
1999
9199
9919
9991
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i just did all this to prove the formula works.
it does.
for larger numbers of digits it would have been much harder to prove.
since it works with 3 and 4 digits, i'm assuming it would work with larger number of digits.
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the general formula appears to be:
number of total digits ! / number of digits that are the same !
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what if you had 9393?
i suspect your answer would be 4!/(2!*2!) = 4*3*2*1/2*2 = 6
let's see if that holds up.
9393
3939
9339
3993
9933
3399
and i think we ran out of possible permutations that are different from each other.
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probably more than you wanted to know, but maybe instructive.
it was for me.
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