Difficult Math Problem #114 - Combinations

[800guy]

A group of 8 friends want 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


from diff math doc, ans coming when some ppl respond with explanations

[Tame the CAT]

I got B

The numerator - there are 8 total people, so 8!

The denominator - There are 4 teams of 2

The set up

(8!)/(2!*2!*2!*2!) = 2520


Another way to look at it

A B C D E F G H
1 1 2 2 3 3 4 4

The letters represent the tennis players and the numbers below it represent the groupings. The total number of players (eight) gets factorized, for lack of a better term and is placed in the numerator. The numbers on the bottom represent the teams. I see two 1s so that is 2! two 2s gets 2! and so on.

What's the answer?

[santy]

Let the players be ABCDEFGH

A can be paired with 7 others
B can be paired with 5 others ( A and 1 more is already paired)
C can be paired with 3 others (A+1, B+1 is already paired)
D can be paired with 1 other remaining (A+1, B+1, C+1 already paired)

So, 7*5*3*1=105.

[Tame the CAT]

^^ Very easy to follow approach. I like that.

I thought my answer seemed far fetched.

[jayhawk2001]

First team can be formed in 8C2 / 4 ways
Second team in 6C2 / 3 ways
Third team in 4C2 /2 ways
last team in 2C2 /1 ways

We have to divide by 4, 3, 2, 1 respectively since there are duplicates
i.e. AB-CD and CD-AB are the same.

So, we have 8C2/4 * 6C2/3 * 4C2/2 * 2C2/1 = 105 ?

[Cybermusings]

What's the official answer?

[vk.neni]

First team can be formed in 8C2 / 4 ways
Second team in 6C2 / 3 ways
Third team in 4C2 /2 ways
last team in 2C2 /1 ways

We have to divide by 4, 3, 2, 1 respectively since there are duplicates
i.e. AB-CD and CD-AB are the same.

So, we have 8C2/4 * 6C2/3 * 4C2/2 * 2C2/1 = 105 ?

Hi Jayhawk2001,
I was approaching the problem the same you did! ie.
8c2 * 6c2 * 4c2 * 2c2. Wouldn't 8c2 = 28?
nCr = n!/(r! * (n-r)!). If so, we'd end up with
28 * 15 * 6 * 1 = 2520.

How do you determine the team-combination duplication?

Do you see something not right?

Thanks
Neni

[jayhawk2001]

Hi Jayhawk2001,
I was approaching the problem the same you did! ie.
8c2 * 6c2 * 4c2 * 2c2. Wouldn't 8c2 = 28?
nCr = n!/(r! * (n-r)!). If so, we'd end up with
28 * 15 * 6 * 1 = 2520.

How do you determine the team-combination duplication?

Do you see something not right?

Thanks
Neni

To simplify, lets take 4 players - ABCD.

Teams that can be formed are

AB CD
AC BD
AD BC

All other possibilities e.g. CD AB are duplicates. So, we have to
divide by 2 here to get rid of the duplicates (hence 4C2/2).

Doing 4C2 already takes care of the ordering i.e. AB vs BA but
since we have 2 teams playing each other, we have to go 1 more
level and prune the duplicates.

[800guy]

oa:

out of 8 people one team can be formed in 8c2 ways.

8c2*6c2*4c2*2c2= 2520.
The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )

[BTGmoderatorRO]

4 teams of 2 people each are to be formed from a group of 8 people.
(i) total number of ways of forming the first team of 2 people is
$$_{8C2=28ways}$$
(ii) when the first team is formed, we have 6 people left to divide into three teams
therefore, total number of ways of forming the second team of 2 people is
$$_{6C2=15ways}$$
(iii) again, we will have 4 people left to divide into 2 teams.
therefore, total number of ways of forming the third team of 2 people is
$$_{6C2=6ways}$$
(iv) we now have 2 people left, making the last team.
Number of ways of forming the fourth team of 2 people is just $$_{2C2=1way}$$.
Therefore, total number of ways of forming the four teams is = 28*15*6*1 =2520ways

[Scott@TargetTestPrep]

A group of 8 friends want 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


from diff math doc, ans coming when some ppl respond with explanations

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

[GMATGuruNY]

A group of 8 friends want 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 person selected can be paired with 7 different people, giving us 7 possible pairs.
8-2 = 6 people left.
The person selected can be paired with 5 different people, giving us 5 possible pairs.
6-2 = 4 people left.
The next person selected person can be paired with 3 different people, giving us 3 possible pairs.
4-2 = 2 people left, giving us 1 more possible 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 want 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


from diff math doc, ans coming when some ppl respond with explanations

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

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Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
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- 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