Archit143 wrote:

Hi

I have a doubt in the question.

6 students can be arranged into 2 groups of 3 each. so ie 2!

Now these 3 students in each group can be arranged in 3! ways in each group so it gives 3!*3!

So total ways in which the group can be arranged = 2!*3!*3! = 72

Can someone help...where did i go wrong

Archit

Responding to a pm:

Two groups are not distinct so you don't have 2!. You did not name the groups as GroupA and GroupB.

(A, B, C) and (D, E, F) split is the same as (D, E, F) and (A, B, C) split.

Also, you do not have to arrange the 3 students in 3! ways. You just have to group them, make a team - not make them stand in a line in a particular sequence.

In fact, you can use the opposite method to understand how to get the answer.

Arrange all 6 in a line in 6! ways.

First 3 is the first group and next 3 is the second group. But guess what, the groups are not distinct so divide by 2!.

Also, since the students needn't be arranged, divide by 3! for each group.

You get 6!/(2!*3!*3!) = 10 (that's how you get the formula)

Also, I have discussed grouping here using different methods:

http://www.veritasprep.com/blog/2011/11 ... ke-groups/ _________________

Karishma

Veritas Prep | GMAT Instructor

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