10
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think that this time you are horribly right. $\endgroup$ – drhab Apr 13 '17 at 16:58
  • 1
    $\begingroup$ Your solution is correct. $\endgroup$ – Soroush khoubyarian Apr 13 '17 at 17:02
6
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.