9. All numbers from 1 to 10000000000 are written in a row. Which numbers will be more - those in which the digit 1 appears, or those in which 1 does not appear?
9. First solution. Let's calculate the number of numbers in our sequence that do not contain the digit 1. Add the number 0 (zero) at the very beginning of this sequence and omit the last number 10000000000; we will get a sequence of $10^{10}$ numbers, containing one more number, in the notation of which the digit 1 does not appear, than the original sequence. Let's agree to prepend as many zeros to all numbers in the obtained sequence that are less than ten digits, so that they become ten-digit. The new sequence will consist of $10^{10}$ ten-digit numbers, starting from 0000000000 and ending with 9999999999. If a number in this sequence does not contain the digit 1, then this means that the first digit in this number is one of the nine digits 0, $2,3,4,5,6,7,8,9$. The second digit of the number must also be one of the same nine digits; combining 9 possible values of the first digit with 9 values of the second, we get a total of $9^{2}$ different possibilities for the first pair of digits. Similarly, for the first three digits of our number, we get $9^{3}$ different possibilities, for the first four digits $9^{4}$ different possibilities, and so on; finally, for the first ten digits, we get $9^{10}$ different possibilities. This means that in the sequence of numbers from 0000000000 to 9999999999, there are $9^{10}$ different numbers that do not contain the digit 1. Therefore, in the sequence of numbers from 1 to 10000000000, such numbers will be

$$
9^{10}-1=3486784401-1=3486784400
$$

and the numbers that contain the digit 1 will be

$$
10000000000-3486784400=6513215600 .
$$

Thus, the numbers that contain the digit 1 will be more.

Second solution. Let's calculate the number of numbers in our sequence that contain the digit 1. Let's agree that a decade means ten consecutive positive integers, starting from a number ending in zero and ending with a number ending in 9; a hundred means a hundred consecutive integers, starting from a number ending in two zeros and ending with a number ending in two nines; a thousand means a thousand consecutive integers, starting from a number ending in three zeros and ending with a number ending in three nines, and so on. To make the first decade, the first hundred, the first thousand, and so on, complete, we will prepend the number 0 to the beginning of our sequence of numbers.

In the first decade, there is a single number that contains the digit 1--this is the number 1. The same number of numbers containing the digit 1 will be included in any other decade of the first hundred, except the second--the second decade (numbers from 10 to 19) consists entirely of numbers containing 1.

Therefore, the first hundred contains

$$
9 \cdot 1+10
$$

numbers that contain the digit 1.

Exactly the same number of numbers that contain the digit 1 are contained in any other hundred of the first thousand, except the second--the second hundred consists entirely of numbers containing 1. Therefore, the first thousand contains

$$
9(9 \cdot 1+10)+100=9^{2}+9 \cdot 10+10^{2}
$$

numbers that contain the digit 1.

In each thousand of the first ten thousand, except the second, there will be the same number of numbers that contain the digit 1, and the second thousand will consist entirely of such numbers. Thus, among the first 10000 numbers, there will be

$$
9\left(9^{2}+9 \cdot 10+10^{2}\right)+1000=9^{3}+9^{2} \cdot 10+9 \cdot 10^{2}+10^{3}
$$

numbers that contain the digit 1.

Continuing to reason in the same way, we can easily show that among the first 10000000000 numbers from 0 to 9999999999, there are

$$
9^{9}+9^{8} \cdot 10+9^{7} \cdot 10^{2}+\ldots+9 \cdot 10^{8}+10^{9}
$$

numbers that contain the digit 1. This sum is easily calculated using the formula for the sum of the terms of a geometric progression:

$$
\begin{aligned}
& 9^{9}+9^{8} \cdot 10+9^{7} \cdot 10^{2}+\ldots+9 \cdot 10^{8}+10^{9}= \\
&= \frac{10^{9} \cdot \frac{10}{9}-9^{9}}{\frac{10}{9}-1}=\frac{\frac{10^{10}}{9}-9^{9}}{\frac{1}{9}}=10^{10}-9^{10}
\end{aligned}
$$

In the sequence from 1 to 10000000000, the numbers containing 1 will be one more, i.e.,

$$
10^{10}-9^{10}+1=6513215600
$$

This result, naturally, coincides with the one obtained in the first solution. It shows that in this sequence, the numbers containing the digit 1 will be more than the numbers that do not contain 1.