# How many six-digit numbers can be formed using the digits $0$ to $9$, where exactly one digit is repeated only once?

How many six-digit numbers can be formed using the digits $$0$$ to $$9$$, where exactly one digit is repeated only once?

For example, some of those numbers can be:
123451
123453
156786
205470

In these $$4$$ numbers, I have only given the possible numbers that have the repeated digit in the last place. But the repeated digit can be in any place like:

898564
432467

So what is the quantity of all possible numbers?

For me:

We know that the 1st digit can't be $$0$$, so $$9$$ possibilities for the first digit. We have $$5$$ remaining places, where one place belongs to the repeated digit and the other $$4$$ places belong to $$4$$ other digits. We also know that the repeated digit can be any of the $$5$$ other digits within the number, and the repeated digit can be in any other place, so $$5$$ places for the repeated digit to go. (Except if it's a $$0$$, then it only has $$4$$ places.)

I'm stumped.

• Divide into cases according to whether the repeated digits is $0$ or not. You may need to separately consider the case in which $0$ isn't one of the five digits at all (in any case, that's an easy scenario to count).
– lulu
Jun 4, 2023 at 17:19
• I believe you have a typo in "..a number with 6 different digits, where one of the 5 digits is repeated only once..." Pl. edit to avoid confusion. Jun 4, 2023 at 19:17
• Welcome to MathSE. Please write a title that is specific to the problem. This MathJax tutorial explains how to typeset mathematics on this site. Jun 5, 2023 at 11:22

Using recommendation of @lulu, lets separate our problem according to existence of zero such that

• Zero repeats two times:$$\binom{5}{2}.9.8.7.6=30240$$

where $$\binom{5}{2}$$ means selecting positions for zeros except for the leading digit

• Zero repeats only once:$$\binom{5}{1}\binom{9}{1}\binom{5}{2}.8.7.6=151200$$

where $$\binom{5}{1}$$ means selecting positions for zero except for the leading digit. $$\binom{9}{1}$$ means selecting repeating digit and $$\binom{5}{2}$$ means selecting position for repeating digits

• No Zero: $$\binom{6}{2}\binom{9}{1}8.7.6.5=226800$$

$$\binom{6}{2}$$ means selecting position for repeating digits and $$\binom{9}{1}$$ means selecting repeating digit

$$30240+151200+226800=408240$$

Hint Temporarily disregard the prohibition on leading $$0$$s. Notice that for any such number, we can increase each of the digits (modulo $$10$$) to produce another such number, and doing this $$10$$ times (and no fewer) returns us to our original number and hence partitions the set of all such numbers into groups of $$10$$. For example, $$001234$$ is a member of the block $$\{001234, 112345, \ldots, 889012, 990123\}$$ Evidently exactly one number in each block---hence exactly $$\frac{1}{10}$$ of all of the numbers---begins with a $$0$$.

So, we might as well determine the number of possibilities disregarding the complicating restriction on leading $$0$$s, after which the desired count is $$\frac{9}{10}$$ that figure.

Disregarding the condition on leading $$0$$s, there are $${6 \choose 2} = 15$$ unordered pairs of positions for the repeated digit, then $$10$$ choices for the repeated digit, $$9$$ choices for the first unrepeated digit, ..., and $$6$$ choices for the last digit, that is, $${}_{10} \mathrm{P}_5 = \frac{10!}{(10 - 5)!}$$ choices for the $$5$$ digits, for a total of $${6 \choose 2} \cdot{}_{10} \mathrm{P}_5 = 453\,600$$ numbers. Applying the reasoning in the hint gives a final count of $$\frac{9}{10}{6 \choose 2} \cdot{}_{10} \mathrm{P}_5 = \frac{9}{10} \cdot 453\,600 = 408\,240 .$$

This is just a simpler exposition of Travis Willse answer which I am upvoting.

Any from the $$10$$ digits chosen, whether as a single or a double, has an equal probability of being the leading digit. Thus required answer is simply

$$\small\text{P(non-0 leading digit)(choose double)(place double)(choose, permute remaining)}$$

$$= \frac9{10}\cdot\binom{10}5\binom62\binom94*4! = 408 240$$

• Sorry I came late to this question, engrossed in French Open ! Let it signal the beginning of a new era ! Jun 6, 2023 at 9:33