I'd go with combinations here: 6C2 = 15.

its definitely a combination question

choose 2 people out of 6 = 6C2 = 15

choose 2 people out of remaining 4 = 4C2 = 6

chose 2 people out of remaining 2 = 2C2 = 1

total ways = 15*6*1 = 90 ways


OA?

Morgoth wrote:its definitely a combination question

choose 2 people out of 6 = 6C2 = 15

choose 2 people out of remaining 4 = 4C2 = 6

chose 2 people out of remaining 2 = 2C2 = 1

total ways = 15*6*1 = 90 ways


OA?

I am sorry Morgoth! Let's take any one person of six, this could be paired with any one of the remaining five persons in 5 ways; once it's done, take any one from the remaining 4 standbyes, this could be paired with any one of the remaining three persons in 3 ways; we are now left with 2 persons only, which is by itself a pair! Now count the various number of ways: isn't that 5 * 3 * 1 = 15, or shall we go DanaJ's way?

Though some people have found the correct answer, I don't think any of these methods would work if the problem asked to split six people into two groups of three.

Morgoth wrote:its definitely a combination question

choose 2 people out of 6 = 6C2 = 15

choose 2 people out of remaining 4 = 4C2 = 6

chose 2 people out of remaining 2 = 2C2 = 1

total ways = 15*6*1 = 90 ways


OA?

Close!

What you haven't factored in is that you've counted the same possibilities multiple times.

For example, under your method you counted:

AB CD EF

but you've also counted

CD EF AB

which are actually the same pairing.

So, we need to factor out the duplicates. Since there are 3 pairs, we need to divide through by 3!, which gives us:

(6C2 * 4C2 * 2C2)/3! = 90/6 = 15

If we were dividing into two groups of 3, we'd have:

(6C3 * 3C3)/2! = 20/2 = 10

OA is 15..
thanks