I will use the fundamental principle of counting to solve this question.
Given the set of numbers, $D = \{2, 2, 3, 3, 3, 4, 4, 4, 4\}$.
Here, count(2's) = 2, count(3's) = 3, count(4's) = 4.
We are required to construct 4-digit numbers greater than 3000. For easy visualisation, we will use dashes on paper, _ _ _ _.
Case 1: Thousandth's place is 3.
Set, $D' = \{2, 2, 3, 3, 4, 4, 4, 4\}$.
Rest of the 3 places can be filled in $3 \times 3 \times 3$ ways provided we have 3 counts of the three unique digits, for filling each of the 3 places. But count(2's) = 2, count(3's) = 2 in the new set. Deficiency for each digit is 1. (Impossible numbers - 3222, 3333).
Therefore, ways to fill = $3 \times 3 \times 3 - (1+1) = 25$.
Case 2: Thousandth's place is 4.
Set, $D' = \{2, 2, 3, 3, 3, 4, 4, 4\}$.
Similarly, the deficiency is of only digit 2, which is 1 count. (Impossible number - 4222)
Therefore ways to fill = $3 \times 3 \times 3 - 1 = 26$.
Summing each case up, we get $26 + 25 = 51$.
Hope this helps.
Best Answer
Count all 7 digit numbers that can be formed using the 7 digits, and then subtract the count of those starting with zero: $$\frac{7!}{2!3!}-\frac{6!}{2!3!} $$