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ketkoag: Legendary Member; Posts: 876; Joined: Thu Apr 10, 2008 8:14 am; Thanked: 13 times

General question

by ketkoag » Tue Mar 31, 2009 1:22 am

There are 6 people.They are divided into 3 pairs. How many ways are there to do that?

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Source: Beat The GMAT — Problem Solving | Tue Mar 31, 2009 1:22 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 am; Thanked: 712 times; Followed by:550 members; GMAT Score:770

by DanaJ » Tue Mar 31, 2009 1:37 am

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

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Morgoth: Master | Next Rank: 500 Posts; Posts: 316; Joined: Mon Sep 22, 2008 12:04 am; Thanked: 36 times; Followed by:1 members

by Morgoth » Tue Mar 31, 2009 2:31 am

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?

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Tue Mar 31, 2009 5:51 am

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?

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relic: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Tue Mar 31, 2009 5:28 am; Thanked: 3 times; GMAT Score:780

by relic » Tue Mar 31, 2009 6:34 am

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.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Tue Mar 31, 2009 11:48 am

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