Solution:

If Zhenya had no restrictions, he could eat the pieces in any order, and he would have $9!=362880$ ways.

Not suitable are the ways in which the central cell is eaten before any of the four adjacent to it. Since for each of these five cells, there is an equal number of options where it is earlier than the others, every fifth option is not suitable. This leaves $\frac{4}{5} \cdot 9!=290304$.

## Open Mathematics Olympiad for School Students

March 11, 2018

## 9th grade  2nd variant  Solutions
8. (4 points) For February 23, the boy Zhenya was given a chocolate bar of size $3 \times 3$, on each piece of which a unique picture is drawn, with each picture appearing only once. In each move, Zhenya can bite off one piece that has no more than three common sides with other pieces that have not yet been eaten. In how many ways can Zhenya eat his chocolate bar?

Answer: $\frac{4}{5} \cdot 9!=290304$.