My maths book tells me that if I want to divide 12 people into 3 groups containing 2,3 and 7 people the following logic applies:
The two people of the first group are chosen in [HTML]<sup>12</sup>C<sub>2</sub>[/HTML] ways i.e. 12!/10!2!
The three people of the second group are chosen in [HTML]<sup>10</sup>C<sub>3</sub>[/HTML] ways i.e. 10!/7!3!
and the third group contains the remainder which can be chosen in only 1 way. So we have:
12! 10! / 10! 2! 7! 3! = 12! / 2!3!7! = 7920 ways.
Then it gives a general principle supporting this logic. "The number of ways of dividing (p+q+r) unlike things into 3 groups containing p, q and r things respectively is: (p+q+r)! / p!q!r!
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?