Team Selection Test

The answer is $2(2018!{)}^2$. We consider the problem for an $n\times n$ board. Let $r_i$ be the number of red squares in the $i$-th row for $i=1,\dots ,n$ and $c_j$ be the number of red squares in the $j$-th column for $j=1,\dots ,n$. Since $r_i\ne r_j$ for every $i\ne j$, we see that $\{r_1,\dots ,r_n\}=\{0,1,\dots ,n\}\setminus \{a\}$ for some number $a$. Similarly, we have $\{c_1,\dots ,c_n\}=\{0,1,\dots ,n\}\setminus \{b\}$ for some number $b$. On the other hand, the number of red squares on the board is both ${\sum }_{i=1}^n{r}_i$ and ${\sum }_{j=1}^n{c}_j$. Thus, we get $a=b$.

Note that if $r_i=0$ for some $i$, then $c_j\le n-1$ for every $j$ and hence, $a=n$. Moreover, if $r_i=n$ for some $i$, then $c_j\ge 1$ for every $j$ and hence, $a=0$. Therefore, the missing number is either $0$ or $n$. If $a=0$, then $r_i=c_j=n$ for some $i$ and $j$ and removing the $i$-th row and $j$-th column gives us the problem for $(n-1)\times (n-1)$ board with the missing number $n-1$. Similarly, If $a=n$, then $r_i=c_j=0$ for some $i$ and $j$ and removing the $i$-th row and $j$-th column gives us the problem for $(n-1)\times (n-1)$ board with the missing number $0$. Therefore, at the beginning we can decide on the missing number in $2$ ways but later on it is uniquely determined. We can choose a row and a column to remove in $n\cdot n$ ways. Finally, the answer is $$2\cdot n\cdot n\cdot (n-1)\cdot (n-1)\cdots 2\cdot 2\cdot 1\cdot 1=2(n!{)}^2.$$