Jill is dividing her ten-person class into two teams of equal size for a basketball game. If no one will sit out, how many different match-ups between the two teams are possible?
A. 10
B. 25
C. 126
D. 252
E. 630
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- Brent@GMATPrepNow
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Let's say we have Team Blue and Team Red. Notice that once we select 5 people to be on Team Blue, the other 5 people must automatically be on Team Red.vinay1983 wrote:Jill is dividing her ten-person class into two teams of equal size for a basketball game. If no one will sit out, how many different match-ups between the two teams are possible?
A. 10
B. 25
C. 126
D. 252
E. 630
In how many ways can we select 5 people to be on Team Blue?
Since the order in which we select the 5 people does not matter, we can use combinations.
We can select 5 people from 10 people in 10C5 ways (= 252 ways)
So, there are 252 different ways to select 5 people to be on Team Blue and 5 people to be on Team Red?
IMPORTANT: Notice that Team Blue = {A,B,C,D,E} and Team Red = {F,G,H,I,J} is EXACTLY THE SAME as team Blue = {F,G,H,I,J} and Team Red = {A,B,C,D,E}.
In fact, we have inadvertently counted each possible configuration TWICE.
So, to get the correct answer, we must take 252 and divide by 2 to get 126
Answer: C
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Brent
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It's easy. You only have to do the Selection. If we select one of the two groups, the other group is selected by default.
Therefore number of ways of selecting 5 people from 10 = C(10, 5) = 252
Therefore number of ways of selecting 5 people from 10 = C(10, 5) = 252
- theCodeToGMAT
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It can't be 252.. Brent has beautifully explained the reason in his solution post,which is above your post.visheshchadha wrote:It's easy. You only have to do the Selection. If we select one of the two groups, the other group is selected by default.
Therefore number of ways of selecting 5 people from 10 = C(10, 5) = 252
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