Since they are 4 digit combinations rather than 4-digit numbers,
they do not have to begin with something other than 0.
Since they are combinations rather than permutations, the order
they are listed in does not matter.
You may have meant something else, but this is the answer to
the way it is stated.
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282303391 or, in order
012233389.
one 0, one 1, two 2's, three 3's, one 8, one 9.
Case 1: All the digits are different.
Choose 4 from these 6 digits: 0,1,2,3,8,9
6C4 = 15
[FYI, Those 15 combinations of 4 digits, not permutations
and not 4-digit numbers, are
{0,1,2,3}, {0,1,2,8}, {0,1,2,9}, {0,1,3,8}, {0,1,3,9},
{0,1,8,9}, {0,2,3,8}, {0,2,3,9}, {0,2,8,9}, {0,3,8,9},
{1,2,3,8}, {1,2,3,9}, {1,2,8,9}, {1,3,8,9}, {2,3,8,9}]
Case 2: There are exactly 2 identical digits.
Choose the digit that we have 2 of in 2 ways, as 2 or 3.
Choose two other digits in 5C2 ways.
That's 2*5C2 = 20 ways
Case 3: There is only one combination that has 2 pairs of
identical digits, {2,2,3,3}
That's 1.
Case 4: 3 identical digits.
Choose the identical digit only 1 way, as 3.
Choose the 4th digit 5 ways 0,1,2,8, or 9
That's 1*5 = 5 ways,
[FYI, they are {0,3,3,3}, {1,3,3,3}, {2,3,3,3}, {3,3,3,8}, {3,3,3,9}
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Grand total: 15+20+1+5 = 41
Edwin
