# numbers which contain the digit $1$

The number of numbers between $$1$$ to $$10^{10}$$ which contain the digit $$1$$ is

what i try:we have to formed total number between $$1$$ to $$10^{10}$$ from $$0,1,2,3,4,5,6,7,8,9$$

$$\bullet\;$$ Single digit number which contain $$1$$ is $$1$$

$$\bullet\;$$ Double digit numbers which contain $$1$$ is $$17$$

But this is very lengthy ways

How do i solve it Help me please

• Consider how many don't contain a $1$ instead, perhaps? And use leading zeros, because that is easier (think $0000000001$ rather than $1$). Jan 22, 2020 at 8:58

Edited shorter answer : Any number with at most ten digits does not have $$1$$ as a digit, if and only if , when represented with leading zeros, it contains in ten possible blanks, any digit from $$0 \to 9$$ except $$1$$. Depending upon the number of leading zeros you would get the number of digits of course.
Therefore, that means that the answer is $$10^{10} - 9^{10}$$, realizing that $$10^{10}$$ contains a $$1$$ as a digit so does not count.
• That seems a bit longwinded. Any number up to $10^{10}-1$ can be represented as a 10-digit number, if necessary with leading zeros. So the number without a 1 is just $9^{10}$. Subtract from $10^{10}$ and then think about whether the OP included $10^{10}$ itself. Jan 22, 2020 at 9:31