emmak 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
A really good question and there is a high chance that a lot of people will mark 252 as the answer.
While grouping people or any item, we always need to take care of double counting. Also, whenever one sees options which are multiple of each other ( in this case 126 and 252),we should always double check our answers to make sure we have not made a mistake of double counting.
The best way to find out if one is making a mistake or not is to take small numbers and check it quickly. Let's say instead of 10, there were 2 people (A and B). In how many ways can you make two teams of equal size?
The answer is simple, right? It's 1. A in one team and B in the other. But if we use the formula, we will get 2C1 = 2, which is not correct, because of double counting, (A and B) and (B and A) have been considered different, which is not correct. Hence, to get the correct answer, we need to divide 2C1 by 2, which would give us 1.
In this question also, we need to do the same thing 10C5 would include a lot of repetitive cases, and we can get rid of them by dividing it by 2. Thus, the correct answer would be 10C5/2, which is Option C.
Regards,
Saquib
e-GMATQuant Expert