1st team = 8C2 = 28 ways
2nd team = 6C2 = 15 ways
3rd team = 4C2 = 6 ways
4th team = 2C2 = 1 way
The number of ways in which the group can be divided
= (28*15*6*1)/4! = [spoiler]105 ways (E)[/spoiler]
(Divided by 4! since the order of the teams do not matter in this case).
Hoping this is correct. Please let me know if not. Thanks!
PnC
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cypherskull
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neelgandham
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Total number of ways of selecting the first team = 8C2
Total number of ways of selecting the second team = 6C2
Total number of ways of selecting the third team = 4C2
Total number of ways of selecting the fourth team = 2C2
Total number of ways of selecting 4 teams = 8C2*6C2*4C2*2C2. But the question asks the total number of distinct ways. So the above value must be divided with 4!
Let A B C D E F G H be the friends, then
AB CD EF GH is the same combination as
AB EF CD GH, which is the same combination as
AB EF GH CD, which is the same combination as
AB GH EF CD, which is the same combination as
AB CD GH EF, which is the same combination as
AB GH CD EF, which is the same combination as
CD AB EF GH, which is the same combination as
CD AB GH EF, which is the same combination as
CD EF AB GH, which is the same combination as
CD EF GH AB, which is the same combination as
CD GH AB EF, which is the same combination as
CD GH EF AB, which is the same combination as
EF AB CD GH, which is the same combination as
EF AB GH CD, which is the same combination as
EF CD AB GH, which is the same combination as
EF CD GH AB, which is the same combination as
EF GH CD AB, which is the same combination as
EF GH AB CD, which is the same combination as
GH AB CD EF, which is the same combination as
GH AB EF CD, which is the same combination as
GH CD AB EF, which is the same combination as
GH CD EF AB, which is the same combination as
GH EF AB CD, which is the same combination as
GH EF CD AB.
Total 24 similar combinations
Answer = 8C2 * 6C2 * 4C2 * 2C2 /24 = 2520/24 = 105
Total number of ways of selecting the second team = 6C2
Total number of ways of selecting the third team = 4C2
Total number of ways of selecting the fourth team = 2C2
Total number of ways of selecting 4 teams = 8C2*6C2*4C2*2C2. But the question asks the total number of distinct ways. So the above value must be divided with 4!
Let A B C D E F G H be the friends, then
AB CD EF GH is the same combination as
AB EF CD GH, which is the same combination as
AB EF GH CD, which is the same combination as
AB GH EF CD, which is the same combination as
AB CD GH EF, which is the same combination as
AB GH CD EF, which is the same combination as
CD AB EF GH, which is the same combination as
CD AB GH EF, which is the same combination as
CD EF AB GH, which is the same combination as
CD EF GH AB, which is the same combination as
CD GH AB EF, which is the same combination as
CD GH EF AB, which is the same combination as
EF AB CD GH, which is the same combination as
EF AB GH CD, which is the same combination as
EF CD AB GH, which is the same combination as
EF CD GH AB, which is the same combination as
EF GH CD AB, which is the same combination as
EF GH AB CD, which is the same combination as
GH AB CD EF, which is the same combination as
GH AB EF CD, which is the same combination as
GH CD AB EF, which is the same combination as
GH CD EF AB, which is the same combination as
GH EF AB CD, which is the same combination as
GH EF CD AB.
Total 24 similar combinations
Answer = 8C2 * 6C2 * 4C2 * 2C2 /24 = 2520/24 = 105
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
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Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
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neelgandham
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Another approach:
Select one person. He/She can be paired with 7 different people(from the remaining lot) giving us 7 pairs. Now the total number of people remaining is 6
Select another person (#3). He/She can be paired with 5 different people(from the remaining 5) giving us 5 different pairs. Now the total number of people remaining is 4
Select another person (#5). He/She can be paired with 3 different people(from the remaining 3) giving us 3 different pairs. Now the total number of people remaining is 2
The last couple will not have a chance to choose.
Total number of pairs = 7*5*3* = 105
Select one person. He/She can be paired with 7 different people(from the remaining lot) giving us 7 pairs. Now the total number of people remaining is 6
Select another person (#3). He/She can be paired with 5 different people(from the remaining 5) giving us 5 different pairs. Now the total number of people remaining is 4
Select another person (#5). He/She can be paired with 3 different people(from the remaining 3) giving us 3 different pairs. Now the total number of people remaining is 2
The last couple will not have a chance to choose.
Total number of pairs = 7*5*3* = 105
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
