In an eight-digit number, each of its digits (except the last one) is greater than the digit following it. How many such numbers are there?

(I. Jančigová)
The so-called popped numbers can be constructed by removing two out of ten available digits and arranging the remaining eight in descending order. This is the same as removing two digits from the ten-digit number 9876543210:

- If we remove 9 as the first digit, then there are nine options left for which digit to remove as the second.
- If we remove 8 as the first digit, then there are eight options left for which digit to remove as the second (removing the pair 8 and 9 is included in the previous case).
- If we remove 7 as the first digit, then there are seven options left for which digit to remove as the second (removing the pairs 7, 8 and 7, 9 is included in the previous cases).
- Similarly, we proceed until the end...
- If we remove 1 as the first digit, then there is only one option left for which digit to remove as the second, and that is 0.

The total number of possibilities, i.e., the number of all numbers that meet the conditions, is

$$
9+8+7+6+5+4+3+2+1=45 \text {. }
$$

Another solution. The number of ways to remove two digits from ten can be determined as follows:

The first digit can be chosen in ten ways from ten. The second digit can be chosen in nine ways from the remaining nine digits. This gives a total of $10 \cdot 9=90$ ways to choose two digits from ten with regard to the order of selection. However, this order does not matter to us - what matters is which digits we choose, not the order in which we choose them. Therefore, the total number of possibilities is half:

$$
\frac{1}{2} \cdot 10 \cdot 9=45
$$

Note. The previous double expression of the same result can be generalized as follows:

$$
(n-1)+(n-2)+\cdots+2+1=\frac{1}{2} \cdot n \cdot(n-1) .
$$

This number represents the number of all pairs that can be formed from $n$ elements (without regard to the order of selection).