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

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:

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

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8C4 * 4C2 = 70 * 6 = 420

E
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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.

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

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)?

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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First, we select the four players. The number of ways to choose 4 people from 8 is 8C4 = (8 x 7 x 6 x 5) / (4 x 3 x 2) = 2 x 7 x 5 = 70.

Next, we split the four selected players into pairs to form the matchups. Let’s call the individuals A, B, C, and D. There are 3 matchups for these four individuals:

AB vs. CD, AC vs. BD, and AD vs. BC

Since there are 70 ways to select 4 people and each selection has 3 ways of creating matchups, there are a total of 70 x 3 = 210 matchups.

Answer: B
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I don't get this one thing. Is the game being played only between the 4 selected individuals out of 8? Are we not including e,f,g,h??

Each match consists of 4 players.

4 players can be drawn from 8
8!/4!4!= 70

The 4 players can be divided into 3 matchups:
AB CD
Ac bd
Ad bc

Total 3x70=210

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Hi,

I have a more intuitive approach to this. I don't know if it's correct, but it lead to the right result:

There are a total of 8!/(6!)/(2)=28 distinct teams possible to form out of 2 distinct employees, with a total of 8 employees to pick from. Now, if we pick one of these 28 teams, they can NOT play match ups against teams which contain at least one of these two players. In total, we can therefore create 6!/(4!)(2)=15 teams where neither one of the two players picked initially are inside.

That leaves (28)(15)=420, but we have to divide by two, because if (a,b) was a matchup between team a and b, then (a,b) is equal to (b,a).

Choosing a pair out of 8 players = 8C2
Choosing another pair from remaining players = 6C2
Total pairs =8C2*6C2 = 420
One match up has 2 pairs = 420 / 2 = 210