NuMbers between 1000 and 9999

[Paulrgibbs]

I have been trying to solve "How many numbers with one or more digits the same are there between 1000 and 9999 (whole numbers only). I have tried analysing the probability of each digit, but am stuck. I would like a hint rather than a solution! Possible answers given are:
62 x 72, 52 x 72, 52 x 82, 42 x 82, or 42 x 92
Help please!
Paul

 

[HallsofIvy]

I would do it the other way around- find the number of integers with NO repeated digits, then subtract from 10000.

 

[pka]

I have been trying to solve "How many numbers with one or more digits the same are there between 1000 and 9999 (whole numbers only).

There is a problem with the phrase in red. In the test writing if the question means $\displaystyle a\le t\le b$ that is numbers from a to b. On the other hand, $\displaystyle a<t<b$ is written numbers between a and b.

I will use the first meaning. There are $\displaystyle 9000$ whole numbers from 1000 to 9999.
Each is four digits long and zero is not the first digit, $\displaystyle 9\times 10 \times 10 \times 10=9000.$
of those $\displaystyle 9\times 9 \times 8 \times 7=4536$ have no repeated digits.
So how many have at least one repeated digit?

 

[Paulrgibbs]

There is a problem with the phrase in red. In the test writing if the question means $\displaystyle a\le t\le b$ that is numbers from a to b. On the other hand, $\displaystyle a<t<b$ is written numbers between a and b.

I will use the first meaning. There are $\displaystyle 9000$ whole numbers from 1000 to 9999.
Each is four digits long and zero is not the first digit, $\displaystyle 9\times 10 \times 10 \times 10=9000.$
of those $\displaystyle 9\times 9 \times 8 \times 7=4536$ have no repeated digits.
So how many have at least one repeated digit?

Thank you for that. It's obvious now, thanks again

 

[Paulrgibbs]

I would do it the other way around- find the number of integers with NO repeated digits, then subtract from 10000.

Thanks fou your help