2007 AMC 10B Problems/Problem 20

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Problem

A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?

$\textbf{(A) } 100 \qquad\textbf{(B) } 125 \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$

Solution

There are $25$ ways to choose the first square. The four remaining squares in its column and row, and the square you chose exclude nine squares from being chosen next time.

There are $16$ remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.

Finally, the last square has $9$ remaining choices.

The number of ways to choose $3$ squares is $25 \cdot 16 \cdot 9,$ but you must divide by $3!$ since some sets are the same.

\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 6}\]

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions