- #1

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S_12

(1,2)(3,4)(5,6)(7,8)(9,10)(11,12) - 10,395

This stems from

[tex]\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}

[/tex][tex]\frac{1}{6!}[/tex]

The 6! gets rid of redundant combinations. Is this right? I don't have a calculator handy so I'm not sure how to calculate that.