NuMbers between 1000 and 9999

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Paulrgibbs

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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
 
I would do it the other way around- find the number of integers with NO repeated digits, then subtract from 10000.
 
Paulrgibbs said:
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).
Click to expand...
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?
 
pka said:
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?
Click to expand...
Thank you for that. It's obvious now, thanks again
 
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