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How many 4-digit numbers can be formed from the digits of number 23222300?
So, we have 2 zeros, 4 twos and 2 threes. I have tried to consider the last 3 digits of 4-digit number, but it doesn't work as we don't use all the digits given.

Note, leading zeroes are not allowed. We form numbers not strings.

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  • $\begingroup$ Are leading zeroes (as in $0023$) allowed? $\endgroup$ – mlc Oct 23 '16 at 16:59
  • $\begingroup$ No, leading zeroes are not allowed. We form numbers not strings. $\endgroup$ – Armen Gabrielyan Oct 23 '16 at 17:03
  • $\begingroup$ One way to proceed is to form all 4-digit strings (i.e. leading zeroes allowed), then count how many of them aren't actual 4-digit numbers and drop these. (I think this leads to an alternating sum?) No idea if this is the simplest approach, though... $\endgroup$ – Semiclassical Oct 23 '16 at 17:08
  • $\begingroup$ Then we can arrange them however we like, and the order of the original number doesn't matter, besides not having the $0$s in front? $\endgroup$ – Kevin Long Oct 23 '16 at 17:16
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Probably the easiest option in this case is to find $4$-digit numbers using digits $\{0,2,3\}$ and then work out which are not possible.

So the unconstrained count of such numbers is $2\times 3\times 3\times 3 = 54$. From this we can remove $8$ options with too many $3$s: $\{3333,2333, 3X33, 33X3, 333X\}$ with each $X$ being $0$ or $2$. Similarly we can remove the $2$ options with too many zeros, $\{2000,3000\}$. Note that we cannot simultaneously have too many $3$s and too many $0$s.

Thus leaving $54-10=44$ options.

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