I have got this problem that says : Given digits $2,2,3,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed.

The options are: $50,51,52,54.$

Is there any way I can logically solve the problem instead of manually counting ?

Also, Almost same problem has been discussed here but I am not satisfied with the answers.


Clearly the only allowed numbers have to have either a $4$ or a $3$ in the thousands place.

If it's a $4$, there are three possible digits to put in the remaining three positions. So the number of possibilities here are $3\times 3\times 3 - 1$ ($-1$ to account for the number $4222$ which is not possible).

Similarly if $3$ is in the thousands place, there are $3\times 3 \times 3-2$ possibilties ($3333$ and $3222$ being excluded).

So the total is $27+27-1-2 = 51$.

  • $\begingroup$ thanks a lot.nice explanation.got it. $\endgroup$ – learner Jun 5 '13 at 5:52

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