The digits $3, 4, 5, 7$ and 9 were used to form all possible three-digit numbers such that each digit appears in each number at most once.

Determine the number of such numbers and their total sum.

(M. Volfová)
When creating three-digit numbers with the given properties, the following procedure can be followed:

The first place can be occupied by any of the five given digits, i.e., 5 possibilities. For each of these placements, the second place can be occupied by any of the four remaining digits, i.e., a total of $5 \cdot 4=$ = 20 possibilities. For each of these placements, the third place can be occupied by any of the three remaining digits, i.e., a total of 5 $4 \cdot 3=60$ possibilities. Thus, the number of three-digit numbers created this way is 60.

To determine the total sum of all these numbers, it is sufficient to realize that each of the five digits appears twelve times in each place $(5 \cdot 12=60)$. The sum of the digits in a selected place in all considered numbers is equal to

$$
12 \cdot(3+4+5+7+9)=12 \cdot 28=336 .
$$

According to the respective place, this value appears once, ten times, and a hundred times in the total sum. Therefore, the sum of all considered numbers is

$$
336 \cdot(1+10+100)=336 \cdot 111=37296 .
$$

Evaluation. 2 points for the number of numbers; 2 points for the total sum; 2 points for the quality, or completeness of the commentary.