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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.