5. Given six cards, on which the digits $1,2,4,5,8$ and a comma are written. From them, all possible numbers are formed (each card must be used exactly once, the comma cannot stand at the beginning or at the end of the number). What is the arithmetic mean of all such numbers?

(M. V. Karluкова)
Solution. First, let's assume there is no comma, that is, we find the sum of integers from the given digits. There are 5! = 120 ways to order the digits. When summing them, each of the five possible digits will appear 4! times in each place value, so the sum of the digits in the units place is $4! \cdot (1+2+4+5+8) = 24 \cdot 20 = 480$, and in the other place values, it is 10, 100, 1000, 10000 times more. Therefore, the sum of the numbers is $S = 480 \cdot 11111$.

However, we are summing not integers but numbers with different positions of the decimal point. A decimal point after four (three, two, one) digits does not change the number of numbers (there are 120 in each case), but reduces their sum by 10, 100, 1000, 10000 times, respectively. Therefore, the total number of numbers is $120 \cdot 4 = 480$, and their sum is $S \cdot 0.1111 = 480 \cdot 11111 \cdot 0.1111$. Thus, the arithmetic mean is $11111 \cdot 0.1111 = 1234.4321$ (which is easy to calculate by column).

Answer: 1234.4321.

Criteria. 2 points - the number of numbers is found.

-1 point for an arithmetic error in the final step, -3 points for an arithmetic error in the calculation in another place.