A set of 16 square blocks is arranged into a 4 x 4 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.
First square can be chosen in $16$ ways then leaving out the squares in the row and column of the square chosen, we are left with $9$ squares, then the second square can be chosen in $9$ ways. Similarly, $3^{rd}$ square can be chosen in $4$ ways . So there are $16\times 9\times 4$ ways. But we don't want the arrangement among the squares then the required answer is $\frac{16\times 9\times 4}{3!}=96$
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