# Numbers of $5$ digits that have at least one of the digits repeated more than one time?

How can I count the numbers of $5$ digits such that at least one of the digits appears more than one time?

My thoughts are:
I count all the possible numbers of $5$ digits: $10^5 = 100000$. Then, I subtract the numbers that don't have repeated digits, which I calculate this way: $10*9*8*7*6$ $= 30240$. Thus, I have $100000 - 30240 = 69760$ numbers that have at least one digit repeated more than one time.

Is this correct?

• If it is 5-digit numbers, isn't the total $10^5$? Commented Mar 2, 2014 at 0:51
• My bad. Fixed it. Commented Mar 2, 2014 at 0:53
• Actually, my bad, in that I was also trying to hint that you need to eliminate the numbers in {$0,1,2,..,9999$} from that total. Commented Mar 2, 2014 at 0:56
• Are you allowing 0 as a leading digit? Commented Mar 2, 2014 at 0:58
• Yes, I am allowing 0 as a leading digit. Commented Mar 2, 2014 at 1:00

No digit is ever repeated for $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6=30240$ numbers.
The total number of $5$ digit numbers is $10^5=100000$.

So the number of $5$ digits numbers with at least $1$ digit repeating more than once is: $$10^5 - 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 69760$$ So, yes you are correct. NOW I noticed I have just repeated your correct arguments. Down vote for it to disappear! :-)

• I am allowing $0$ as a leading digit, so it would be $10⋅9⋅8⋅7⋅6 = 30240$ and the total number of $5$ digit numbers would be $10^5$. Commented Mar 2, 2014 at 1:03

You're doing it wrong.

Total no. of $5$ digit numbers that are possible are $9⋅10⋅10⋅10⋅10$ because $0$ can't be kept on the first place, else it will end up being a $4$ digit number. $= 90000$

Similarly, when any digit is not repeated, it would be $9⋅9⋅8⋅7⋅6$ as one option in less for the first digit here too. $= 27216$

Hence, the answer is $90000-27216 = 62784$

Helpful question: In how many ways can we make or arrange five digits so that there is no repetition?

We know that there are 10 possible digits to choose from for the first digit, 0 - 10. However, we can't have the first digit be a 0, else we get a four-digit number. So, we're down to 9. For each of the 9 digits, there are 9 digits to choose from for the second digit to avoid repetition and so on.

This gives $$9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27,216$$ ways.

Actual question: How many five-digit numbers have at least one digit that occurs more than once?

We also know that there are $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90,000$$ possible five-digit numbers. At this point, it should be clear why the first digit can only be selected in 9 ways.

In order to get the number of five-digit numbers that have at least one digit repeated, we simply subtract the number of possible numbers without repeated digits from the total number of five-digit numbers, which gives us $$90,000 - 27,216 = 62,784$$

My answer is probably redundant at this point. Nevertheless, I'll just leave it here. :)