From the given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many 4-digit numbers…

When digits from a limited multiset (some digits repeated only a fixed number of times) are arranged to form a number satisfying a size condition, fix the leading digit(s) that satisfy the condition, then count arrangements of the remaining positions from the leftover supply -- subtracting only the arrangements that would need more copies of a digit than remain available.

1. The multiset {2, 2, 3, 3, 3, 4, 4, 4, 4} supplies digit 2 twice, digit 3 three times, and digit 4 four times (9 tiles in total).

2. A 4-digit number greater than 3000 must begin with 3 or 4, since a number beginning with 2 is at most 2444 (< 3000) and no 0 or 1 tile is available.

3. First digit 3: the leftover supply becomes {2:2, 3:2, 4:4}. The remaining 3 positions can each independently be 2, 3, or 4, giving 33 = 27 raw sequences; the sequences that repeat one digit all three times need 3 copies of it, which exceeds the leftover supply of both digit 2 (only 2 left) and digit 3 (only 2 left) -- removing these 2 sequences leaves 27 - 2 = 25 valid numbers.

4. First digit 4: the leftover supply becomes {2:2, 3:3, 4:3}. Again 33 = 27 raw sequences; only the all-2 sequence exceeds its leftover supply (needs 3, only 2 left). The all-3 and all-4 sequences each need exactly 3 copies, and the leftover supply still has exactly 3 of each, so both remain valid. Removing only 1 sequence leaves 27 - 1 = 26 valid numbers.

5. Total = 25 + 26 = 51.

An independent check by directly listing every ordered draw of 4 tiles from the 9 available and counting the distinct resulting integers above 3000 also gives exactly 51 (25 beginning with 3, 26 beginning with 4), confirming the count.

So 51 four-digit numbers greater than 3000 can be formed.