# 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.

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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 (2,2,3,4,4,4,4). So any place can have either 2 or 3 or 4, but (222,333,332,323,233,334,343,433) is not possible, because we have just one 3's and two 2's

So total = 3*3*3 - 8 = 19

If first place is 4 - so now we have to choose 3 digits from (2,2,3,3,4,4,4). 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*3*3 - 2 = 25

 There's a total of $\binom{8}{4}$ ways to choose four digits, and that includes repetition, so the total will be even less that that. $1260$ is too high. – ladaghini Aug 26 '12 at 7:21 @ladaghini: please see the edited answer. I previously applied a wrong approach. – Rahul Taneja Aug 26 '12 at 7:54
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.