Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed? Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.  
 A: 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.
A: You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3, 3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333)$ is not possible, because we have just two $3$'s  and two $2$'s
So total = $3 \times 3\times 3 - 2 = 25$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3, 3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222)$ is not possible, because we have just two $2$'s
So total = $3 \times 3\times 3 - 1 = 26$
So, total choices = $25+26 = 51$, which is your answer.
A: It is a matter of detailed counting.  You have to start with a $3$ or $4$ to be greater than $3000$.  If you start with a $4$, you have three choices for each other space, except you can't have $222$ or $333$, so there are $25$.  If you start with a $3$ it is harder as you have a different number of each digit left, but it is the same idea.
