gmat_guy666 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)

A. 8
B. 10
C. 16
D. 20
E. 24

OA B

By the official answer, it appears that the two groups are not unique. That is, there's no group 1 and group 2. There are just two groups of boys.


HOWEVER, for my solution let's PRETEND that there IS a group 1 and a group 2.
Now select 3 boys to be in group 1.
Since the order in which we select the boys doesn't matter, we can use combinations.
We can select 3 boys from 6 boys in 6C3 ways (= 20 ways).
By default, the remaining 3 boys are automatically placed in group 2.

Aside: if anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Of course there is no group 1 and group 2, so our answer of 20 isn't correct, because the above solution counts each grouping twice.
For example, selecting boys A, B, and C for group 1 (leaving D, E, and F in group 2) is the SAME as selecting boys D, E, and F for group 1 (leaving A, B, and C in group 2), because there is no distinction among the two groups.

Since the initial solution counted each grouping TWICE, the correct answer is 20/2 = [spoiler]10 = B[/spoiler]

Cheers,
Brent

gmat_guy666 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)

A. 8
B. 10
C. 16
D. 20
E. 24

Here's a solution that doesn't require us to divide by 2 at the end.

Let's say the 6 boys are A, B, C, D, E and F
We can create two groups of 3 by combining boy A with two other boys (and the remaining 3 boys are automatically placed in the other group).

So, in how many ways can we assign 2 boys to be the same group as boy A?
There are 5 boys remaining, and we must select 2 to join boy A.
Since the order in which we select the boys doesn't matter, we can use combinations.
We can select 2 boys from 5 boys in 5C2 ways (= 10 ways).

Answer: B

Cheers,
Brent