A group of 8 friends want to play doubles tennis.

[VJesus12]

A group of 8 friends wants to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

The OA is the option E.

What is the formula that helps me here? Combinations or permutations? I am confused. <i class="em em-confused"></i>

[GMATGuruNY]

A group of 8 friends wants to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Since there are 8 people, the first friend selected can be paired with 7 different people, yielding 7 possible pairs.
Since 6 people remain, the next friend selected can be paired with 5 different people, yielding 5 possible pairs.
Since 4 people remain, the next friend selected can be paired with 3 different people, yielding 3 possible pairs.
The 2 remaining friends must form the 1 last pair.
To combine the options in blue, we multiply:
7*5*3*1 = 105.

The correct answer is E.

[Brent@GMATPrepNow]

A group of 8 friends wants to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Let the 8 people be: A, B, C, D, E, F, G, and H

Take the task of creating the teams and break it into stages.

Stage 1: Select a partner for person A
There are 7 people to choose from, so we can complete stage 1 in 7 ways

ASIDE: There are now 6 people remaining. Each time we pair up two people (as we did in stage 1), we'll next focus on the remaining person who comes first ALPHABETICALLY.
For example, if we paired A with B in stage 1, the remaining people are C, D, E, F, G and H. So, in the next stage, we'll a partner for person C.
Likewise, if we paired A with E in stage 1, the remaining people are B, C, D, F, G and H. So, in the next stage, we'll a partner for person B.
And so on...

Stage 2: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 5 people remaining, so we can complete this stage in 5 ways.

Stage 3: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 3 people remaining, so we can complete this stage in 3 ways.

Stage 4: Select a partner for the remaining person who comes first ALPHABETICALLY
There is 1 person remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create 4 pairings) in (7)(5)(3)(1) ways (= 105 ways)

Answer: E
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

[Jeff@TargetTestPrep]

A group of 8 friends wants to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

The first team can be selected in 8C2 = (8 x 7)/2! = 28 ways.

The next team can be selected in 6C2 = (6 x 5)/2! = 15 ways.

The next team can be selected in 4C2 = (4 x 3)/2! = 6 ways.

The final team can be selected in 2C2 = 1 way.

However, since ORDER OF THE TEAMS DOES NOT MATTER, we need to divide the total number of ways to select the teams by 4! since we have 4 different teams. So we have:

(28 x 15 x 6)/4! = 2520/24 = 105

Answer: E