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An office comprised of eight employees is planning to have a foosball

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An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 31 Jul 2017, 23:57
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An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?

(A) 70

(B) 210

(C) 280

(D) 336

(E) 420
[Reveal] Spoiler: OA

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Re: An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 01 Aug 2017, 00:00
Bunuel wrote:
An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?

(A) 70

(B) 210

(C) 280

(D) 336

(E) 420


8C4 * 4C2 = 70 * 6 = 420

E
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Re: An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 12 Aug 2017, 23:52
Don't know if my reasoning is right, but here's my take:
Every match is done with 4 people, so we have to find out in how many ways we can split up those 8 in groups of 4: 8C4.
Now that we have 2 groups of 4, separate them in pairs: 4C2. Thing is, do we have pair A, B, etc? I may be wrong, but I don't we do, so:

(8C4*4C2)/2! = 210.

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Re: An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 13 Aug 2017, 08:21
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Bunuel wrote:
An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?

(A) 70

(B) 210

(C) 280

(D) 336

(E) 420


Take the task of creating a matchup and break it into stages.

Stage 1: Select 4 employees
Since the order in which we select the employees does not matter, we can use combinations.
We can select 4 employees from 8 employees in 8C5 ways (70 ways)
So, we can complete stage 1 in 70 ways

Stage 2: Divide the 4 selected employees into 2 teams
Let's say the 4 selected employees are A, B, C, D
A nice way to determine the number of ways to divide the 4 employees into 2 teams is to find a partner for one person.
For example, let's find a partner for employee A.
NOTE: once we choose a partner for employee A then, by default, the remaining two two employees will be paired together.
In how many ways can we select a partner for employee A? Well, A can be paired with B, C or D
So, we can complete stage 2 in 3 ways

ASIDE: The 3 pairings are:
AB vs CD
AC vs BD
AD vs BC

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a matchup) in (70)(3) ways (= 210 ways)

Answer:
[Reveal] Spoiler:
B


Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS





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Re: An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 26 Aug 2017, 15:12
First, recognize that picking 4 people from a group of 8 (to create a single match up) is simply: 8*7*6*5, aka 8!/(8-4)!

Then, realize the number of redundant groupings:
1. Team A can be (1,2) or (2,1)
2. Team B can be (3,4) or (4,3)
3. Team A vs Team B is another redundancy (A,B) or (B,A)

So, you end up with (8*7*6*5)/(2!*2!*2!)

At this point, I like to break down 8*7*6*5 to primes: (2^4*7*3*5)

This, divided by 2^3 leads to: 7*3*2*5 = 21*10 = 210

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Re: An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 26 Aug 2017, 19:58
GMATPrepNow wrote:
Bunuel wrote:
An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?

(A) 70

(B) 210

(C) 280

(D) 336

(E) 420


Take the task of creating a matchup and break it into stages.

Stage 1: Select 4 employees
Since the order in which we select the employees does not matter, we can use combinations.
We can select 4 employees from 8 employees in 8C5 ways (70 ways)
So, we can complete stage 1 in 70 ways

Stage 2: Divide the 4 selected employees into 2 teams
Let's say the 4 selected employees are A, B, C, D
A nice way to determine the number of ways to divide the 4 employees into 2 teams is to find a partner for one person.
For example, let's find a partner for employee A.
NOTE: once we choose a partner for employee A then, by default, the remaining two two employees will be paired together.
In how many ways can we select a partner for employee A? Well, A can be paired with B, C or D
So, we can complete stage 2 in 3 ways

ASIDE: The 3 pairings are:
AB vs CD
AC vs BD
AD vs BC

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a matchup) in (70)(3) ways (= 210 ways)

Answer:
[Reveal] Spoiler:
B


Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS






Thanks for the great explanation. But I have a doubt. For stage two you have considered only the first 4 employees a,b,c,d and not the rest e,f,g,h.
Does it mean, if we get the no. of ways for just 4 employees and multiply it by the stage 1's 70, then we automatically consider the rest four(e,f,g,h)?

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An office comprised of eight employees is planning to have a foosball [#permalink]

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New post 26 Aug 2017, 20:35
Bunuel wrote:
An office comprised of eight employees is planning to have a foosball game. A matchup consists of four players, split into pairs. If any employee can be paired up with any other employee, then how many unique matchups result?

(A) 70

(B) 210

(C) 280

(D) 336

(E) 420


Hi..

the Q would be a simple combinations BUT for pairs..

If you understand that when you take out TWO out of FOUR, the OTHER 2 automatically make a PAIR, you have your answer.

Choose 4 out of 8 in 8C4.
Now choose two out of these 4 in 4C2 ways, but divide by 2! because of the reason mentioned above..
EXAMPLE
4 person ABCD.. way to choose two is 4C2 but this includes AB and CD as 2 different cases..
However when you choose AB, you are automically making other two CD as a pair.


ans = \(8C4*4C2*\frac{1}{2!}=\frac{8!}{4!*4!}*\frac{4!}{2!2!}*\frac{1}{2}=\frac{8*7*6*5}{8}=210\)
B
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An office comprised of eight employees is planning to have a foosball   [#permalink] 26 Aug 2017, 20:35
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