Jill is dividing her ten-person class into two teams of eq

Question

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

Bunuel · Accepted Answer

There should be 5 people in each group. We can divide a group of 10 people into 2 teams of 5 in \frac{C^5_{10}*C^5_5}{2!}=126 ways (dividing by 2! because the order of the groups doesn't matter). Answer: C. For more on this check: a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html in-how-many-different-ways-can-a-group-of-9-people-be-85993.html probability-88685.html probability-85993.html combination-55369.html sub-committee-86346.html Hope it helps.

EMPOWERgmatRichC · Answer

Hi All,

This question tests your knowledge of the Combination Formula, but it comes with a rare "twist" that most people don't realize. Here's a bit more information on that "twist":

With 10 players, the process of figuring out how many groups of 5 can be formed is pretty straight-forward....

10C5 = 10!/(5!5!) = 256 possible groups of 5

Once forming that first group of 5, the remaining 5 players would all be placed on the second team by default.

The 'twist' is that the two teams of 5 can "show up" in either order:

For example, if we call the two teams of 5 players: A,B,C,D,E and F,G,H,I,J

ABCDE vs. FGHIJ

is the SAME match-up as....

FGHIJ vs. ABCDE

So we are NOT allowed to count that matchup twice. This means we have to divide the 256 by 2.

Final Answer:  C

GMAT assassins aren't born, they're made,
Rich

LalaB · Answer

here is a formula - https://gmatclub.com/forum/a-group-of-8-friends-want-to-play-doubles-tennis-how-many-55369.html#p689312 
 The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is -
(mn)!/(n!)^m*m!

10! /((5!)^2*2!) =126

bluetrain · Answer

good question.

When we pick 5 people out of 10 people(10C5), we can make each side of team simultaneously.
And We have to devide by 2 to avoid duplication.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: (C)

This is a trickier spin on a basic combinatorics problem. Begin by asking how many groups of 5 can be created from a pool of 10 candidates. Order does not matter, so this is a combinations problem. Use the combinations formula: n!/(k!)(n-k)! That answer would be 10!/  = 252. But (D) is a trap answer. The problem does not ask how many team arrangements are possible; it asks how many match-ups are possible. Each match-up consists of a team of five opposing the remaining five people. Therefore, each match-up involves two of the 252 teams. To find the number of match-ups, divide 252 in half for a total of 126. The correct answer is (C).

HiLine · Answer

I love questions like these which test your problem-solving skill rather than pure math. If you don't sit back to make sure your approach is on point before diving into the numbers, you'll likely pick the wrong answer.

EgmatQuantExpert · Answer

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-GMAT Quant Expert

AkshdeepS · Answer

Good example used to check double counting. It cleared my mind quite nicely and I will check many trap answers this way. Thanks

Henry S. Hudson Jr. · Answer

One good idea is to start with a smaller case. Suppose there are 4 teams. A,B,C, and D

The possibilities of combinations taken two at a time are

AB 
 AC 
 AD

BC 
 BD

CD

I have color coated the "pairs." If AB is selected they must face off against CD. If CD is selected they must face off against AB. So those two combinations cannot be counted twice. This leads us to reason we should take a combination of 10 things taken 5 at a time and then divide by 2 to account for matching "pairs."

{(10)(9)(8)(7)(6)}/{(5)(4)(3)(2)(1)(2)}=126

bumpbot · Answer

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Jill is dividing her ten-person class into two teams of eq

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Jill is dividing her ten-person class into two teams of eq

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