number theory

As I read the question, the answer should be 6!, and not (6!)^2 (which is the same as 6^2 * 5^2 * 4^2 * 3^2 * 2^2 * 1^2). If you arrive at the answer (6!)^2, then you're assuming the order of the selection matters; that is, you're considering the selection (6, 11, 16, 21, 26, 31) to be different from the selection (31, 26, 21, 16, 11, 6), and you're counting them twice. Of course, there are 6! orders in which you could sequence each set of numbers, so the answer (6!)^2 is too large by a factor of 6!. Divide by 6!, and you'll get the right answer if order does not matter: 6!.

It's easier just to fix a particular order - we have six choices we can make from the first row, five from the second, etc, which gives the answer 6!.