Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?

Answer: $720$.

Attempt at deriving the answer:

We have two cases: four digit numbers and five digit numbers.

Five digit numbers:

Let $x \in \{2, 3, 4, 5\}$. If the first position in a five digit number is not $x$, then there are $5$ possibilities for this position as there are four values for $x$ and $0$ is inadmissable. The rest of the four positions will have the various permutations of four values of $x$. There are $5 \times 4!$ such numbers. If the non-$x$ value is in the second position, then there are $4$ ways to choose an $x$-value for the first position, $6$ integers for the second position and $3!$ permutations for the rest of the positions. There are $4\times 6 \times 3!$ such numbers. If the non-$x$ value is in the third position, then there are $\binom 42$ ways to choose two $x$-values, $2!$ ways to permute them and $6$ integers for the third position meaning there are $6 \times 2 \times 6$ such numbers. When the non-$x$ value is in the fourth position, there are $4 \times 3! \times 6$ such numbers. Finally, if non-$x$ is in the fifth position, there are $4! \times 6$ such numbers.

Four digit numbers:

We just need to permute the number $2345$. There are $4!$ such permutations.

Thus the number of numbers with the given restrictions is $5\times 4! + 4\times 3!\times 6 + 2\times 6 \times 6 + 4 \times 3! \times 6 + 6 \times 4! + 4! = 648$.

What did I forget to take into account? Thanks.

  • $\begingroup$ Very nicely thought of answer! But I think you missed one (several) cases in the 5 integer case: There can be 6 options for the non $\;2,3,4,5\;$ digit, as far as its position is not the first one (we count zero here), and there 5 options for that digit if its position is the first one...and etc. $\endgroup$ – DonAntonio Aug 5 '20 at 15:16

To answer the question you asked: In the case where the non-$x$ value is in the third position, you missed permuting the fourth and fifth digits of the number, so that term should be $6\cdot 2\cdot 6\cdot 2$ (rather than $6\cdot 2\cdot 6$).


Write $0$ at the front of four digits number so we always have five digits number. The fifth number is one of $(0,1,6,7,8,9)$.

$6\times 5! =720$

  • 1
    $\begingroup$ Great, concise solution. $\endgroup$ – Andrew Chin Aug 5 '20 at 15:19

Four digit numbers are accounted for by the $4!=\underline{24}$ ways to arrange the four digits $2,3,4,5$.
Five digit numbers not containing $0$ are accounted for by $5\times5!=\underline{600}$ (arranging the four digits $2,3,4,5$ then one from $\{1,6,7,8,9\}$).
Five digit numbers containing $0$ are accounted for by the $5!-4!=\underline{96}$ ways to arrange the five digits $0,2,3,4,5$ while fixing $0$ to the second to fifth digit.

In total, we have $24+600+96=\boxed{720}$ as desired.


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