Arrange in ascending order those numbers that contain each of the digits $1,2, \ldots, 9$ exactly once. What number stands in the 100,000th position?
Among the numbers in question, it is sufficient to examine those that do not contain the digit 0, as the number of exactly nine-digit suitable numbers is $9!=362$ 880, meaning that all suitable numbers with at least ten digits will have a larger serial number.

If we fix the digit with the highest place value, then by writing the remaining eight digits in all possible orders in the next eight places, we get a total of $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=8!=40320$ different nine-digit numbers. If the first place is a 1, we get the first 40320 numbers; if it is a 2, we get the next $40320$ numbers, that is, from the 40321st to the 80640th; if it is a 3, we get the numbers from the 80641st to the 120960th. It is clear that the first digit of the 100,000th number is 3. We then need to determine which of the eight-digit numbers, each containing the digits 1, 2, 4, 5, 6, 7, 8, 9 exactly once, is the $100000-80640=19360$th in size.

If we now fix the first digit of an eight-digit number, we get $7!=5040$ different numbers for each of the remaining seven digits in all possible orders. $19360=3 \cdot 5040+4240$, so by selecting the fourth digit in sequence from the digits that can be in the second place, we can get the 19,360th eight-digit number. This digit is now the 5. Moving forward, we need to find the 4240th number among the seven-digit numbers formed from the digits 1, 2, 4, 6, 7, 8, 9.

By fixing the first digit of a seven-digit number, we get $6!=720$ different numbers. $4240=5 \cdot 720+640$, so we need to select the sixth digit, the 8, for the third place. Now we need to find the 640th number among the six-digit numbers formed from the digits 1, 2, 4, 6, 7, 9.

By fixing the first digit of a six-digit number, we get $5!=120$ different numbers. $640=5 \cdot 120+40$, so the 640th number will be among those starting with the sixth available digit, the 9. We need to find the 40th number among the five-digit numbers formed from the digits 1, 2, 4, 6, 7.

By fixing the first digit of a five-digit number, we get $4!=24$ different numbers. $40=1 \cdot 24+16$, so the second available digit, the 2, will be the next digit. We then need to find the 16th number among the four-digit numbers formed from the digits 1, 4, 6, 7. Moving forward, $16=2 \cdot 3!+4$, so we need to select the third digit from the above four digits, and the 16th number will be among the three-digit numbers formed from the digits 1, 4, 7. $4=2 \cdot 2!$, so the first digit of the 4th number will be the second digit from the above three digits, the 4, and the last two digits will be 7 and 1.

The sought number is therefore 358926471.

Note. If we denote the set of numbers that each contain exactly one of the first nine positive integers by $H$, then essentially we need to count how many $H$-belonging numbers are smaller than a given $a=a_{9} a_{8} \ldots a_{1} \in H$. These numbers can be grouped according to the first place value where a smaller digit appears in them compared to $a$. If this is the $i$-th place, then we can write all the digits smaller than $a_{i}$ that have not yet been used on the larger place values than $i$ on this place, and the remaining $i-1$ places can have all the $(i-1)!$ possible orders of the remaining $(i-1)$ numbers.

If $k_{i}$ denotes the number of digits smaller than $a_{i}$ that appear after $a_{i}$ in the sequence $a_{9} a_{8} \ldots a_{1}$, then the number of $H$-belonging numbers smaller than $a$ is

$$
k_{9} \cdot 8!+k_{8} \cdot 7!+\ldots+k_{3} \cdot 2!+k_{2} \cdot 1!\quad\left(k_{1}=0\right)
$$

We need the $H$-belonging number $a$ for which the above value is exactly 99,999. It is known that every natural number can be written in the above form, in "factorial number system," that is, for every natural number $t$ there is a non-negative integer $n$ such that

$$
t=f_{n} \cdot n!+f_{n-1} \cdot(n-1)!+\ldots+f_{2} \cdot 2!+f_{2} \cdot 2!+f_{1} \cdot 1!,
$$

If we also assume that $f_{i} \leqq i$, then the above representation is unique. Since in (1) $k_{i} \leqq i-1$, because there are $(i-1)$ numbers in total after $a_{i}$ in the sequence, the coefficients in the factorial representation of 99999 are exactly the $k_{i}$ numbers. Essentially, the solution involves writing 100000 in the form (2).

$$
99999=2 \cdot 8!+3 \cdot 7!+5 \cdot 6!+5 \cdot 5!+1 \cdot 4!+2 \cdot 3!+1 \cdot 2!+1 \cdot 1!
$$

that is, $k_{9}=2, k_{8}=3, k_{7}=5, k_{6}=5, k_{5}=1, k_{4}=2, k_{3}=1$ and $k_{2}=1$. The permutation $a_{9} a_{8} \ldots a_{1}$ of the first nine positive integers will be the sought number, in which for every $i$ greater than 1, exactly $k_{i}$ numbers smaller than $a_{i}$ appear after $a_{i}$ in the sequence.

It is clear that if the digits with indices greater than $i$ have already been obtained, then $a_{i}$ will be the $\left(k_{i}+1\right)$-th largest among the remaining $i$ numbers. Thus, we sequentially obtain the sought digits: $a_{9}=3, a_{8}=5, a_{7}=8, a_{6}=9, a_{5}=2$, $a_{4}=6, a_{3}=4, a_{2}=7$. The only remaining digit, the 1, will be $a_{1}$, so the sought number is 358926471.