Arrangements question.

If I have a 5 by 6 matrix with 30 elements, $a_{i,j}$, in how many ways can I select 3 elements $a^1_{i,j}, a^2_{i,j}, a^3_{i,j}$ such that none of the $i$ are the same and none of the j are the same(none in the same column or row).

I think the answer is $\frac{30*20*12}{3!} = 1200$, since their are 30 ways to pick for the first element, 20 for the second, 12 for the third and each arrangement is then counted 6 ways. But I'm not sure about this, and I have a tendency to get these kinds of problems horribly wrong.

• I think that this time you are horribly right. – drhab Apr 13 '17 at 16:58
• Your solution is correct. – Soroush khoubyarian Apr 13 '17 at 17:02

Pick the three horizontal and three vertical coordinates in $\binom{5}{3}\binom{6}{3}$ ways, then permute in $6$ ways.