Originally Posted by

**s_ingram** A couple of pages on my maths book asks: how many ways can 12 people be arranged into groups if there are to be 2 groups of six people? I would have used the same technique i.e. 12! / 6! 6! = 924

But the books provides the following logic:

The first group of 6 can be selected in [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways and the second group of 6 is then automatically made up of those not in the first group, so there is only one way of selecting the second group, but since the [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways include the case when a1,a2,a3...a6 are in the first group and a7, a8... a12 are in the second group and vice versa we divide by 2 (i.e. by the number of ways the two groups can be permutated amongst themselves) and the result is 12! / 6! 6! 2 = 462.

So my question is this: how do you divide into groups? using method 1 or method 2?