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Jill is dividing her tenperson class into two teams of eq [#permalink]
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Updated on: 28 Feb 2013, 06:14
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Jill is dividing her tenperson class into two teams of equal size for a basketball game. If no one will sit out, how many different matchups between the two teams are possible? A. 10 B. 25 C. 126 D. 252 E. 630
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Originally posted by emmak on 28 Feb 2013, 05:25.
Last edited by Bunuel on 28 Feb 2013, 06:14, edited 1 time in total.
Edited the question.



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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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28 Feb 2013, 06:29



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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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02 Mar 2013, 11:47
here is a formula  http://gmatclub.com/forum/agroupof8friendswanttoplaydoublestennishowmany55369.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
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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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23 Apr 2015, 15:42
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 straightforward.... 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 matchup 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: GMAT assassins aren't born, they're made, Rich
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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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04 Oct 2017, 20:14
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.



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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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30 Dec 2017, 12:56
emmak wrote: Jill is dividing her tenperson class into two teams of equal size for a basketball game. If no one will sit out, how many different matchups between the two teams are possible?
A. 10 B. 25 C. 126 D. 252 E. 630 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!)(nk)! That answer would be 10!/[(5!)(5!)] = 252. But (D) is a trap answer. The problem does not ask how many team arrangements are possible; it asks how many matchups are possible. Each matchup consists of a team of five opposing the remaining five people. Therefore, each matchup involves two of the 252 teams. To find the number of matchups, divide 252 in half for a total of 126. The correct answer is (C).
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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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30 Dec 2017, 16:09
I love questions like these which test your problemsolving 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.



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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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31 Dec 2017, 08:49
emmak wrote: Jill is dividing her tenperson class into two teams of equal size for a basketball game. If no one will sit out, how many different matchups 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 eGMATQuant Expert
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Re: Jill is dividing her tenperson class into two teams of eq [#permalink]
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31 Dec 2017, 10:05
EgmatQuantExpert wrote: emmak wrote: Jill is dividing her tenperson class into two teams of equal size for a basketball game. If no one will sit out, how many different matchups 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 eGMATQuant Expert Good example used to check double counting. It cleared my mind quite nicely and I will check many trap answers this way. Thanks
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Jill is dividing her tenperson class into two teams of eq [#permalink]
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31 Dec 2017, 13:35
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




Jill is dividing her tenperson class into two teams of eq
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