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

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.

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

Cheers,
Brent