The 16th Japanese Mathematical Olympiad - The First Round

Denote by $a_{ij}$ the entry that lies in the $i$-th column and the $j$-th row. The given conditions yields that each $1$, $2$, $3$ and $4$ appear exactly once in each column. We may assume without loss of generality that $a_{1j}=j$ for $j=1,2,3,4$. We would multiply the answer by $4!$ afterwards. We may also assume that $a_{22}=a_{33}=1$. We would multiply the answer by $3\times 2$ afterwards. If $a_{21},a_{31}\ne 4$, then $a_{23}=a_{32}=4$. We can choose which of $a_{24}$ and $a_{34}$ to be $2$, and the rest entries would be determined uniquely. So we have two ways. Otherwise, if $a_{21}=4$, there are only one way to fill the other squares. The case $a_{31}=4$ is similar. So we have two ways. Therefore, the answer is $4!\times 3\times 2\times (2+2)=576$.