## Task 19/75

How many five-digit natural numbers with five different digits have a cross sum of 18?

| 0 1 | 2 | 6 | 9 | 0 | 2 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 2 | 7 | 8 | 0 | 2 |
| 0 | 3 | 5 | 9 | 0 | 2 |
| 0 | 3 | 6 | 8 | 0 | 3 |
| 0 | 4 | 5 | 8 | 1 | 2 |
| 0 | 4 | 6 | 7 | 1 | 2 |
| $0 \quad 2$ | 3 | 4 | 9 | 1 | 2 |

(Note: Four-digit numbers with a leading zero should not be considered as five-digit numbers.)
First, one lists lexicographically (in the order of the natural numbers) all increasing sequences of five digits with the sum of 18:

Each of these 14 sequences can be permuted $5!=120$ times, so that one obtains $14 \cdot 120=1680$ different sequences of five different digits with the sum of 18. When forming the five-digit numbers, all sequences that begin with the digit 0 must be eliminated. These are the first eleven of the above list, where the following four digits can be permuted $4!=24$ times.

Thus, $11 \cdot 24=264$ sequences are eliminated when forming the five-digit numbers. Therefore, there are 1416 five-digit numbers with different digits and a cross sum of 18.