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:

The answer to the above question is given by \(\frac{C^3_6*C^3_3}{2!}\), which equals 10. The division by \(2!\), the factorial of the number of teams, is performed because the order of the groups does not matter.

For example, let's consider a group of 6 boys named A, B, C, D, E, and F. One possible grouping using \(C^3_6*C^3_3\) is {ABC} as the first group and {DEF} as the second group, resulting in two groups: {ABC} and {DEF}. However, \(C^3_6*C^3_3\) will also yield the group {DEF} with {ABC} as the second group, resulting in the same two groups: {ABC} and {DEF}. To avoid counting duplicates, we divide \(C^3_6*C^3_3\) by the factorial of the number of teams, which is 2!.


Answer: B

This concept is also discussed at:
https://gmatclub.com/forum/combinations ... er#p734396
https://gmatclub.com/forum/split-the-gr ... ilit=split
https://gmatclub.com/forum/9-people-and ... to#p788744
https://gmatclub.com/forum/ways-to-divi ... to#p763471

Hope it helps.