How many ways are there to split a group of 6 boys into two

GENERAL RULE:
1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is NOT important is \(\frac{(mn)!}{(n!)^m*m!}\).

BACK TO THE ORIGINAL QUESTION:
In original question as the order is NOT important, we should use second formula, \(mn=6\), \(m=2\) groups \(n=3\) objects (people):
\(\frac{(mn)!}{(n!)^m*m!}=\frac{6!}{(3!)^2*2!}=10\).

This can be done in another way as well: \(\frac{C^3_6*C^3_3}{2!}=10\), we are dividing by \(2!\) as there are 2 groups and order doesn't matter.

For example if we choose with \(C^3_6\) the group {ABC} then the group {DEF} is left and we have two groups {ABC} and {DEF} but then we could choose also {DEF}, so in this case second group would be {ABC}, so we would have the same two groups: {ABC} and {DEF}. So to get rid of such duplications we should divide \(C^3_6*C^3_3\) by factorial of number of groups - 2!.

Answer: B.

This concept is discussed at: combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396

Hope it helps.