On average, how many lottery draws result in one where the five numbers drawn contain all ten digits?
I. solution. The customary, ascending order listing of the drawn numbers in the announcements is not an essential part of the draw. We will ignore the sorting in determining both the favorable cases for our problem and the total number of possible draws, imagining that we write down the five numbers of each draw in the order they are drawn. Let us call these raw draws.

The total number of possible raw draws is $N=90 \cdot 89 \cdot 88 \cdot 87 \cdot 86$, because the first number can be chosen in 90 ways, the second from the remaining 89, thus from the 90 possible starts, $90 \cdot 89$ two-number continuations can be formed, the further drawn numbers then allow 88, 87, and finally 86 further developments, and increase the previous number by the respective factors.

The raw draws with the desired property contain each of the ten digits exactly once. By concatenating such five numbers into a single ten-digit number, the single 0 digit stands in an even position, as it only appears as the second digit in a drawn number. Thus, the writing of the favorable raw draws starts with choosing the position of the 0 digit, for which there are 5 possible starts.

Let's take the 9 as the second. This can be written in any of the 4 remaining even positions, but from the odd positions, only before the 0, because 9 only starts the number 90. Thus, there are 5 positions available for the 9, and the placement of the 0 and 9 digits in the 10 positions is possible in $5 \cdot 5$ ways. The remaining digits can appear as tens and units digits of the lottery numbers without restriction, so for the placement of the single 8, 7, ..., 2 digits, there are respectively $8, 7, \ldots, 2$ possibilities according to the number of currently empty positions, and the 1 is always written in the last remaining empty position. Thus, the formation of the concatenated ten-digit number is possible in

$$
N_{k}=5 \cdot 5 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2
$$

ways, which is the number of the sought raw draws. (The order of placement of the 8, 7, ..., 1 digits is irrelevant.)

The total number of raw draws is $N / N_{k} \approx 5232$ times the number of raw draws containing all ten digits, so on average, 1 draw with the desired property occurs every 5232 draws. (Assuming 1 draw per week, this is about once every 100 years.)

Szücs András (Budapest, Fazekas M. High School III. grade)

II. solution. The raw draws containing all ten digits can also be counted as follows. First, consider those that do not contain the number 90. Select the first (tens place) digits of the numbers. These are chosen from the digits 1, 2, ..., 8. The first can be chosen in 8 ways, for each of its choices, the second in 7, the third in 6, the fourth in 5, and the fifth in 4 ways, so the five first digits can be chosen in $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ ways. The remaining five digits must be placed as second digits. The second digit of the first number can be chosen in 5 ways, for the second number from the remaining digits in 4 ways, for the third, fourth, and fifth in 3, 2, and 1 way respectively, the second digit of the fifth number being the remaining digit. Thus, the number of favorable raw draws not containing 90 is

$$
n_{1}=8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 5 \cdot 4 \cdot 3 \cdot 2
$$

Similarly, if 90 is among the drawn numbers, then the first digits of the remaining 4 numbers can be chosen from the digits $1,2, \ldots, 8$ in $8 \cdot 7 \cdot 6 \cdot 5$ ways, and for each such quartet, the remaining 4 digits can be written as second digits in $4 \cdot 3 \cdot 2$ ways, so the 4 numbers can be chosen in

$$
n_{2}=8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2
$$

ways. To each such choice, the 90 can be added as the first, second, third, fourth, or fifth number, so the total number of favorable raw draws is again

$$
N_{k}=n_{1}+5 n_{2}=8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot(4 \cdot 5+5)=5 \cdot 5 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2
$$

Draschitz Rudolf (Budapest, Landler J. Mechanical Engineering High School III. grade)