Number of ways to form three distinctive items Given a 6 by 5 array, Calculate the number of ways to form a set of three distinct items such that no two of the selected items are in the same row or same column.
What I did was $C(30,1) \cdot C(20,1) \cdot C(12,1)$ however this is not the answer. They get 1200. How?

$1^{st}$ item: you will have $6\times5=30$ choices.
$2^{nd}$ item: you take out the row and column containing the $1^{st}$ chosen item, so you are left with $5\times4=20$ choices.
$3^{nd}$ item: you take out the row and column containing the $2^{nd}$ chosen item, so you are left with $4\times3=12$ choices.
However, note that the order of items doesn't matter (i.e choosing $ABC$ is the same as choosing $CBA$). Hence the desired answer is $(30\times20\times12)\div3!=1200$
