It is currently 15 Jan 2018, 23:58

### 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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139136 [0], given: 12776

How many different ways can a group of 6 people be divided into 3 team [#permalink]

### Show Tags

02 Jun 2016, 03:32
Expert's post
4
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

75% (00:49) correct 25% (01:25) wrong based on 81 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139136 [0], given: 12776

Math Expert
Joined: 02 Aug 2009
Posts: 5511

Kudos [?]: 6398 [1], given: 122

Re: How many different ways can a group of 6 people be divided into 3 team [#permalink]

### Show Tags

02 Jun 2016, 05:50
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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 = $$\frac{90}{3!} = \frac{90}{15} = 6$$
_________________

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 [?]: 6398 [1], given: 122

Manager
Joined: 07 Jan 2015
Posts: 91

Kudos [?]: 25 [0], given: 654

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

02 Jun 2016, 19:45
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.

why not divide by 2!?

Can you explain further?

Kudos [?]: 25 [0], given: 654

Math Expert
Joined: 02 Aug 2009
Posts: 5511

Kudos [?]: 6398 [0], given: 122

Re: How many different ways can a group of 6 people be divided into 3 team [#permalink]

### Show Tags

02 Jun 2016, 20:30
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.

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)!
_________________

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 [?]: 6398 [0], given: 122

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7854

Kudos [?]: 18446 [0], given: 237

Location: Pune, India
Re: How many different ways can a group of 6 people be divided into 3 team [#permalink]

### Show Tags

02 Jun 2016, 22:11
Expert's post
3
This post was
BOOKMARKED
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

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18446 [0], given: 237 Intern Joined: 06 Mar 2015 Posts: 29 Kudos [?]: 7 [0], given: 176 Re: How many different ways can a group of 6 people be divided into 3 team [#permalink] ### Show Tags 03 Jun 2016, 06:00 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? Kudos [?]: 7 [0], given: 176 Math Expert Joined: 02 Aug 2009 Posts: 5511 Kudos [?]: 6398 [0], given: 122 Re: How many different ways can a group of 6 people be divided into 3 team [#permalink] ### Show Tags 03 Jun 2016, 06:28 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.... _________________ 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 [?]: 6398 [0], given: 122 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7854 Kudos [?]: 18446 [0], given: 237 Location: Pune, India Re: How many different ways can a group of 6 people be divided into 3 team [#permalink] ### Show Tags 05 Jun 2016, 20:42 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 18446 [0], given: 237

Manager
Joined: 29 Nov 2011
Posts: 114

Kudos [?]: 24 [0], given: 367

Re: How many different ways can a group of 6 people be divided into 3 team [#permalink]

### Show Tags

05 Jun 2016, 21:41
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

Kudos [?]: 24 [0], given: 367

Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 611

Kudos [?]: 835 [0], given: 59

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

18 Dec 2017, 21:42
Top Contributor
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 = $$\frac{90}{3!} = \frac{90}{15} = 6$$

_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

Kudos [?]: 835 [0], given: 59

Re: How many different ways can a group of 6 people be divided into 3 team   [#permalink] 18 Dec 2017, 21:42
Display posts from previous: Sort by