It is currently 20 Jan 2018, 13:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139592 [0], given: 12794

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

### Show Tags

31 Jul 2017, 22:57
Expert's post
7
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

26% (01:41) correct 74% (01:39) wrong based on 119 sessions

### HideShow timer Statistics

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

_________________

Kudos [?]: 139592 [0], given: 12794

Director
Joined: 18 Aug 2016
Posts: 627

Kudos [?]: 208 [0], given: 167

Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: An office comprised of eight employees is planning to have a foosball [#permalink]

### Show Tags

31 Jul 2017, 23:00
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
_________________

We must try to achieve the best within us

Thanks
Luckisnoexcuse

Kudos [?]: 208 [0], given: 167

Intern
Joined: 21 Sep 2016
Posts: 28

Kudos [?]: 2 [0], given: 261

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

### Show Tags

12 Aug 2017, 22:52
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.

Kudos [?]: 2 [0], given: 261

SVP
Joined: 11 Sep 2015
Posts: 1999

Kudos [?]: 2872 [2], given: 364

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

### Show Tags

13 Aug 2017, 07:21
2
KUDOS
Expert's post
Top Contributor
2
This post was
BOOKMARKED
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

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

[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

_________________

Brent Hanneson – Founder of gmatprepnow.com

Kudos [?]: 2872 [2], given: 364

Intern
Joined: 02 Aug 2017
Posts: 1

Kudos [?]: 0 [0], given: 0

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

### Show Tags

26 Aug 2017, 14:12
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

Kudos [?]: 0 [0], given: 0

Manager
Joined: 27 Jun 2015
Posts: 53

Kudos [?]: 9 [0], given: 52

WE: Information Technology (Computer Software)
Re: An office comprised of eight employees is planning to have a foosball [#permalink]

### Show Tags

26 Aug 2017, 18:58
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

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

[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)?

Kudos [?]: 9 [0], given: 52

Math Expert
Joined: 02 Aug 2009
Posts: 5536

Kudos [?]: 6441 [0], given: 122

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

### Show Tags

26 Aug 2017, 19:35
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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Kudos [?]: 6441 [0], given: 122

An office comprised of eight employees is planning to have a foosball   [#permalink] 26 Aug 2017, 19:35
Display posts from previous: Sort by