GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Oct 2019, 09:29

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

How many different ways can a group of 6 people be divided into 3 team

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 02 Jun 2016, 04:32
2
9
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (01:32) correct 32% (01:59) wrong based on 156 sessions

HideShow timer Statistics

Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7971
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 02 Jun 2016, 06:50
2
1
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)
_________________
Current Student
avatar
B
Joined: 08 Jan 2015
Posts: 77
Location: Thailand
GMAT 1: 540 Q41 V23
GMAT 2: 570 Q44 V24
GMAT 3: 550 Q44 V21
GMAT 4: 660 Q48 V33
GPA: 3.31
WE: Science (Other)
How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 02 Jun 2016, 20:45
chetan2u wrote:
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)


I dont fully understand about the dividing by 3! part.

what about this question?

ten-highschool-boys-gather-at-the-gym-for-a-game-of-basketball-two-te-219483.html

why not divide by 2!?

Can you explain further?
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7971
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 02 Jun 2016, 21:30
1
Aves wrote:
chetan2u wrote:
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)


I dont fully understand about the dividing by 3! part.

what about this question?

ten-highschool-boys-gather-at-the-gym-for-a-game-of-basketball-two-te-219483.html

why not divide by 2!?

Can you explain further?



In the Q mentioned by you, we are dividing in two teams so first a team could be selected in 10C5 or in 5C1..
so two place swhere it can be selected and that is why we divide by 2..

But here we are dividing by 3! because any particular team can be selected in 6C2 or 4C2 or 2C1...

say the members are abcdef...
We select in 6C2 = ab...........4C2 = cd..........2C1= ef....
in another case in 6C2 = ab...........4C2 = ef..........2C1= cd....
in another case in 6C2 = cd...........4C2 = ef..........2C1= ab....
in another case in 6C2 = cd...........4C2 = ab..........2C1= ef....
in another case in 6C2 = ef...........4C2 = ab..........2C1= cd....
in another case in 6C2 = ef...........4C2 = cd..........2C1= ab....

so the distribution is the same BUT it has been calculated as 6 different ways....
so we divide total by 6 or 3! ways..
so we divide by (# of groups)!
_________________
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 02 Jun 2016, 23:11
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36


I like to think of it in this way:

Make everyone stand in a straight line. The number of ways of doing this is 6!.

Now, first two people from the left make team A, next two make team B and last two make team C.
We have divided 6 people in three teams (A, B and C) and allocated spots to each member in his team (first person, second person). But the three teams are not distinctly named teams so you un-arrange by dividing by 3!.
Also each team doesn't have an arrangement of "first person and second person" so you un-arrange them by dividing by 2! for each team.

6!/(3!*2!*2!*2!) = 15

Answer (C)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern
Intern
avatar
Joined: 06 Mar 2015
Posts: 25
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 03 Jun 2016, 07:00
chetan2u wrote:
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)


Hi,

I understand that 6 people need to be divided into teams of 2 each. And as it is a team, the arrangement inside should make no difference at all.

Hence 6C2 - 6*5/2 = 15 teams. I can't understand as to why you are again considering "4C2 ways from the remaining people". I guess am missing something monumental here, but i just can't get it. Is the logic that out of 6 people after forming teams of 2 people, you are still left with 4 people and hence can form teams of 2 again. This is my guess but i can't understand how can you consider that, as once you have divided them into 15 teams, you are left with nothing to re-divide?
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7971
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 03 Jun 2016, 07:28
nishi999 wrote:
chetan2u wrote:
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)


Hi,

I understand that 6 people need to be divided into teams of 2 each. And as it is a team, the arrangement inside should make no difference at all.

Hence 6C2 - 6*5/2 = 15 teams. I can't understand as to why you are again considering "4C2 ways from the remaining people". I guess am missing something monumental here, but i just can't get it. Is the logic that out of 6 people after forming teams of 2 people, you are still left with 4 people and hence can form teams of 2 again. This is my guess but i can't understand how can you consider that, as once you have divided them into 15 teams, you are left with nothing to re-divide?



Hi,

if the Q said that we have to choose two people out of 6, then the answer would be straight 6C2..
But here we are talking of 3 teams of 2 each that is why we select 2 out of 6, then another 2 out of remaining 4 and then take remaining 2 as a team..
so these three 6C2, 4C2 and reamining 2, form a set of 3 teams of 2 players each..

Here since order does not matter, we divide it by 3!, so the answer finally comes same as 6C2....
_________________
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 05 Jun 2016, 21:42
nishi999 wrote:

I understand that 6 people need to be divided into teams of 2 each. And as it is a team, the arrangement inside should make no difference at all.

Hence 6C2 - 6*5/2 = 15 teams. I can't understand as to why you are again considering "4C2 ways from the remaining people". I guess am missing something monumental here, but i just can't get it. Is the logic that out of 6 people after forming teams of 2 people, you are still left with 4 people and hence can form teams of 2 again. This is my guess but i can't understand how can you consider that, as once you have divided them into 15 teams, you are left with nothing to re-divide?



From your question I am guessing that you are confusing combinations with permutations.
6C2 is combination which is "only selection". Out of 6 people you select 2 in 15 ways. You do not arrange anyone.

You have 6 people: A, B, C, D, E, F

You can select two in 15 ways:
1. A, B
2. A, C
3. A, D
...
...
15. E, F

You do nothing with the leftover 4 people. So in the next step, you select 2 more out of the 4 remaining to make another team. Finally, you are left with 2 people which make the third team.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager
Manager
avatar
Joined: 29 Nov 2011
Posts: 91
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 05 Jun 2016, 22:41
1
1
1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is (mn)!/ (n!^ m )* m!

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is (mn)!/ (n!^ m )

Here order is not important so we have to divide and we apply formula 1
Answer is 15 (C)
GMAT Tutor
User avatar
G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 18 Dec 2017, 22:42
Top Contributor
Please update 3! value to 6 and answer to 15
chetan2u wrote:
Bunuel wrote:
How many different ways can a group of 6 people be divided into 3 teams of 2 people each?

(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
...


Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{15} = 6\)

_________________
Intern
Intern
avatar
B
Joined: 31 Aug 2016
Posts: 45
How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 12 May 2018, 07:07
1
The smoothest way to solve this is by thinking it like this:

- We have 6 people (A, B, C, D, E, F) and we want 3 teams. Therefore we will have 3 2-member teams. The final teams will be for example: AB, CD, EF.
- Imagine a 6x6 table with A, B, C, D, E, F horizontally and vertically. Those are the possible 2-member teams.
- You need the half triangle without the mid-boxes. Meaning you don't want AA, BB, CC, DD, EE, FF and you don't want to double count AB and BA etc..
- Therefore 6*6=36. Minus the mid-boxes we have 30. And the bottom half we want is 15.
Intern
Intern
avatar
B
Joined: 29 Apr 2019
Posts: 11
Re: How many different ways can a group of 6 people be divided into 3 team  [#permalink]

Show Tags

New post 01 Jul 2019, 02:05
standyonda wrote:
The smoothest way to solve this is by thinking it like this:

- We have 6 people (A, B, C, D, E, F) and we want 3 teams. Therefore we will have 3 2-member teams. The final teams will be for example: AB, CD, EF.
- Imagine a 6x6 table with A, B, C, D, E, F horizontally and vertically. Those are the possible 2-member teams.
- You need the half triangle without the mid-boxes. Meaning you don't want AA, BB, CC, DD, EE, FF and you don't want to double count AB and BA etc..
- Therefore 6*6=36. Minus the mid-boxes we have 30. And the bottom half we want is 15.


Hey found the way really nice, but when I tried the same way for below mentioned question, I could not get correct answer of 3
Q- In how many ways can 4 people be divided into 2 groups of 2 people each?
Did I miss something ?


size=80]Posted from my mobile device[/size]

Posted from my mobile device
GMAT Club Bot
Re: How many different ways can a group of 6 people be divided into 3 team   [#permalink] 01 Jul 2019, 02:05
Display posts from previous: Sort by

How many different ways can a group of 6 people be divided into 3 team

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne