## Task A-3.2.

How many four-digit natural numbers are there in whose notation there are exactly two different digits, each appearing twice?
## First solution.

The first digit can be determined in 9 ways (any except zero).

1 point

There are 3 ways to determine the position where the same digit appears for the second time. 2 points

The same digit must appear on the remaining two positions, and we can choose this digit in 9 ways since we cannot repeat the first digit of the number.

The total number of such numbers is $9 \cdot 3 \cdot 9=243$.

Note: The last point should be awarded regardless of whether the student writes the result as $9 \cdot 3 \cdot 9$ or $3^{5}$ or 243.