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?
 A: $1^{st}$ 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 $(30\times20\times12)\div3!=1200$
A: i found a two part answer to this question i would like to discuss. Perhaps i am over-thinking it. The answer goes as follows:
There are 5!/3!2! = 10 ways to select the three columns in which the three items will appear. The row of the rightmost selected item can be chosen in any of six ways, the row of the leftmost selected item can then be chosen in any of five ways, and the row of the middle selected item can then be chosen in any of four ways. The answer is therefore 10 ⋅ 6 ⋅ 5 ⋅ 4 = 1200. You can also solve this in the following alternative way. There are 30 ways to select the first item. Because there are 10 squares in the row or column of the first selected item, there are 30 − 10 = 20 ways to select the second item. Because there are 18 squares in the rows or columns of the first and second selected items, there are 30 − 18 = 12 ways to select the third item. The number of permutations of these three items is 3!. The number of ways sought is therefore 30 ⋅ 20 ⋅123! = 1200.
Now the 2nd answer was given earlier in this thread. Notation wise i suppose this is a combination problem n!/k!(n-k)! and we assume that the numerator is left with the highest 3 choices: (n * n-1 * n-2)/3!
But the first answer confounds me. It suggests that the rows are a combination and the columns a permutation or vice versa: 5!/3!2! * 6!/3! or 5!/2! * 6!/3!3! 
What gives? Am i missing something?
Thanks for any help.
A: Debes darle importancia a una de las dos, o la fila o la columna. Cuando quieras darle la importancia a una o a la otra, una va con permutación 3 y la otra con combinación 3, para hallar una especie de coordenada. Quedaría 5C3 * 6P3 o lo mismo 6C3 * 5P3. Esto teniendo en cuenta que las 3 respuestas quedarían en linea distinta y 2 de ellas no se encontrarían, porque una esta en la mitad. Con la permutación ya se eliminarían una de las lineas y también se eliminaría el problema de que se encuentren.
