2007 AMC 10B Problem 20

Three blocks with no two in the same row or column use three different rows and three different columns. So build a choice in three stages.

1. Which rows? Pick $3$ of the $5$ rows. Leaving out $2$ rows is the same as choosing which $3$ to keep, and there are $\frac{5 \times 4}{2} = 10$ ways.
2. Which columns? Same count: $10$ ways.
3. Match them up. Say the rows are $r_1, r_2, r_3$ and the columns are $c_1, c_2, c_3$. The block in row $r_1$ can use any of the $3$ columns, the block in row $r_2$ one of the $2$ left, and row $r_3$ gets the last one: $3 \times 2 \times 1 = 6$ ways.
4. Each set of three blocks comes from exactly one choice of rows, columns and matching, so multiply:
$$10 \times 10 \times 6 = 600.$$

The answer is $\boxed{\textbf{(C)}\ 600}$.

This is like placing three rooks on a chessboard so none attack each other. Splitting the choice into "which rows and columns" and "how they pair up" counts each set of blocks once, so no dividing is needed at the end.

Pick the blocks one at a time. The first can be any of $25$. The second must avoid the first one's row and column, leaving a $4 \times 4$ grid: $16$ choices. The third avoids two rows and two columns, leaving $3 \times 3 = 9$. That gives $25 \times 16 \times 9 = 3600$, but each set of three blocks was counted once for every order you could pick them in, $3 \times 2 \times 1 = 6$ orders. So the answer is $3600 \div 6 = 600$.

A valid set of three blocks uses three distinct rows and three distinct columns, and the set is determined by which rows, which columns, and how they are paired up.

• Choose the $3$ rows: $\binom{5}{3} = 10$ ways.
• Choose the $3$ columns: $\binom{5}{3} = 10$ ways.
• Assign each chosen row to a different chosen column: $3! = 6$ ways.

Every combination of blocks arises from exactly one such choice, so the count is
$$10 \cdot 10 \cdot 6 = 600.$$

The answer is $\boxed{\textbf{(C)}\ 600}$.