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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