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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 = $3 \times 3\times 3 - 8 = 19$

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 = $3 \times 3\times 3 - 2 = 25$

So, total choices = $19+25 = 44$, which is your answer.

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  • $\begingroup$ 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. $\endgroup$ – ladaghini Aug 26 '12 at 7:21
  • $\begingroup$ @ladaghini: please see the edited answer. I previously applied a wrong approach. $\endgroup$ – Rahul Aug 26 '12 at 7:54
  • $\begingroup$ But 44 is not an option. The correct answer is given as 51. I wonder how. $\endgroup$ – Ramit Dec 30 '16 at 4:53
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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.

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  • $\begingroup$ Can you provide your final answer please? $\endgroup$ – Ramit Dec 30 '16 at 4:54
  • $\begingroup$ @Ramit: I agree with the $44$ of the accepted answer. Why do you think it is $51$? $\endgroup$ – Ross Millikan Dec 30 '16 at 5:29
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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.

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