In how many ways can we remove 8 letters from the 16 letters in Figure 1 so that two are missing from each row and each column?

| $A$ | $B$ | $C$ | $D$ |
| :---: | :---: | :---: | :---: |
| $E$ | $F$ | $G$ | $H$ |
| $I$ | $J$ | $K$ | $L$ |
| $M$ | $N$ | $O$ | $P$ |
| $A$ | $B$ | $C$ | $D$ |
| :--- | :--- | :--- | :--- |
| $E$ | $F$ | $G$ | $H$ |
| $I$ | $J$ | $K$ | $L$ |
| $M$ | $N$ | $O$ | $P$ |

Solution I. Each selection always belongs to only one of the following three cases.

1. Case. The 4 letters selected from the 1st and 2nd rows are in 4 different columns. From the 4 letters of the first row, 2 can be selected in $\binom{4}{2}$ ways. After that, 2 letters can be selected from the second row in only one way. From the 3rd row, 2 letters can again be selected in $\binom{4}{2}$ ways, because so far we have taken only one letter from each column. Finally, 2 letters can be selected from the 4th row in only one way. Therefore, in this case, there are $\binom{4}{2}\binom{4}{2}=6^{2}=36$ selections.
2. Case. Among the letters selected from the 1st and 2nd rows, 2 are in the same column. From the first row, letters can be selected in $\binom{4}{2}$ ways, and from the second row, in $2 \cdot 2$ ways. Now there is a column from which we have not yet selected any letters. When we select letters from the 3rd row, we must also select the letter from this column, and we cannot select from the column from which we have already taken 2. Thus, there are 2 selection possibilities in the 3rd row, and 1 in the 4th row. Therefore, in this case, there are $\binom{4}{2} \cdot 4 \cdot 2=48$ ways to select the 8 letters.
3. Case. In the 2nd row, we select the letters below the letters selected from the 1st row.

In this case, the letters selected from the 1st row uniquely determine which letters can be selected from the other rows, so there are 6 different selections possible.

Summarizing: $36+48+6=90$ ways to select 8 letters from the diagram so that 2 are missing from each row and column.

Csikós Zsolt (Budapest, Kalamár J. Secondary School, 8th grade)

Solution II. Cut the square containing the letters into 16 small squares, and count how many ways we can color 8 of them black. We can assume that the top-left one is colored black. In this case, from the other three squares in the first column and the first row, we can freely (and independently) choose one each. We can assume that we choose the first possibility in both cases. Then we either color the second square in the second row or we do not. If we color it, the further coloring is already determined. If we do not color it, we can freely choose one each from the remaining two squares in the second row and the second column, and the situation only becomes determined after that. The total number of possibilities is therefore $2 \cdot 3 \cdot 3(1+2 \cdot 2)=90$.