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Expert replies
by Mission2012 » Thu Aug 15, 2013 7:02 pm
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4

(B) 9

(C) 15

(D) 24

(E) 36
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Source: — Problem Solving |

by ganeshrkamath » Thu Aug 15, 2013 7:55 pm
Mission2012 wrote:How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4

(B) 9

(C) 15

(D) 24

(E) 36
ABCDEF

With AB as one team, the other 2 teams can be (CD,EF), (CE,DF) or (CF, ED).
Similarly we can have 3 teams each with AC, AD, AE and AF.
So total combinations = 3*5 = 15

Choose C
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by GMATGuruNY » Thu Aug 15, 2013 8:04 pm
Mission2012 wrote:How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4

(B) 9

(C) 15

(D) 24

(E) 36
Approach 1:
From the 6 people, the number of ways to choose 2 for the first team = 6C2 = (6*5)/(2*1) = 15.
From the 4 remaining people, the number of ways to choose 2 people for the second team = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining people, the number of ways to choose 2 people for the third team = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
15*6*1.
Since the ORDER of the teams doesn't matter -- AB-CD-EF is the same way of dividing the 6 people as CD-EF-AB -- the result above must be divided by the number of ways to ARRANGE the 3 teams (3!):
(15*6*1)/(3*2*1) = 15.

The correct answer is C.

We could also GRIND IT OUT.
Let the 6 people be A, B, C, D, E and F.
Person A must be paired with one of the 5 other people.
Options: AB, AC, AD, AE, AF.

Groupings that can be combined with the first red pair (AB):
CD-EF
CE-DF
CF-DE.
Total options = 3.

The same reasoning can be applied to all 5 of the red pairs above.
Since there will be 3 options for each of the 5 red pairs, the number of ways to divide the 6 people into pairs = 3*5 = 15.
Last edited by GMATGuruNY on Fri Aug 16, 2013 6:08 am, edited 1 time in total.
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by vipulgoyal » Thu Aug 15, 2013 8:25 pm
let the 6 persons be a,b,c,d,e,f
a can be paired with b,c,d,e,f in 5 differant ways
b can be paired with c,d,e,f in 4 ways ( ab and ba are same so counted only once)
5+4+3+2+1= 15
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by rairavig » Fri Aug 16, 2013 4:00 am
its understood that when ever we make pairs from 6 persons will result 3 teams.
so total possible combinations are 6C2= 15
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by Brent@GMATPrepNow » Fri Aug 16, 2013 5:40 am
rairavig wrote:its understood that when ever we make pairs from 6 persons will result 3 teams.
so total possible combinations are 6C2= 15
Hmm, I'm not sure I follow this solution.
Are you saying that, if we wanted to make pairs from 8 people, the answer would then be 8C2? If so, then this approach is not correct.

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