yufenshi wrote:

How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

8

10

16

20

24

The official answer is B, but I don't understand why in this case we need to divide by 2. It seems to me to be the same questions as: how many ways to choose 3 people out of 6 people. in that case it would be 20.

Thanks!

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 also discussed at:

combinations-problems-95344.html?hilit=dividing%20objects%20order#p734396split-the-group-101813.html?hilit=split9-people-and-combinatorics-101722.html?hilit=divided%20equally%20into#p788744ways-to-divide-99053.html?hilit=divided%20equally%20into#p763471Hope it helps.

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