Permutations..

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Tue Jan 12, 2010 11:46 am

Six students are equally divided into 3 groups, then, the three groups were assigned to three different
topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

kindly help me with the solution stepwise

thanks in advance
apoorva

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Source: Beat The GMAT — Problem Solving | Tue Jan 12, 2010 11:46 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Jan 12, 2010 12:08 pm

apoorva.srivastva wrote:Six students are equally divided into 3 groups, then, the three groups were assigned to three different
topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

Here's my edited response.

Let's say we have 3 topics: Topic A, Topic B and Topic C.
Our goal is to assign 2 people to each topic.
Let's take this task and break it into stages.

Stage 1: Select 2 people for Topic A.
Since the order in which we select these people does not matter (e.g., selecting Joe then Al is the same as selecting Al then Joe), we can use combinations.
There are 6 people, and we must select 2. This can be accomplished in 6C2 ways (15 ways).

Stage 2: Select 2 people for Topic B.
Once again, order does not matter.
There are 4 people remaining, and we must select 2. This can be accomplished in 4C2 ways (6 ways).

Stage 3: Select 2 people for Topic C.
There are only 2 people remaining, so this can be accomplished in 1 way

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus assign 2 people to each topic) in (6)(5)(1) ways ([spoiler]= 90 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Last edited by Brent@GMATPrepNow on Tue Jul 02, 2013 5:12 am, edited 4 times in total.

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Tue Jan 12, 2010 12:22 pm

hey brent thanks for the detailed explanation!!

but the answer is 540!!

it is 90* 3!= 540 :(

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Jan 12, 2010 12:34 pm

apoorva.srivastva wrote:Six students are equally divided into 3 groups, then, the three groups were assigned to three different
topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

kindly help me with the solution stepwise

thanks in advance
apoorva

Brent's approach gives us the number of different teams we can make. We now need to figure out how many different ways we can distribute the 3 topics to those teams.

Let's think of the 3 topics as 3 slots to fill. If we have 3 groups for 3 slots, there are 3! different ways to arrange them.

So, we have 90 possible groups and 6 different ways to assign them to topics, giving us:

90 * 6 = 540 possibilities.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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by Brent@GMATPrepNow » Wed Jan 13, 2010 10:00 am

Stuart Kovinsky wrote:
apoorva.srivastva wrote:Six students are equally divided into 3 groups, then, the three groups were assigned to three different
topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

kindly help me with the solution stepwise

thanks in advance
apoorva
Brent's approach gives us the number of different teams we can make. We now need to figure out how many different ways we can distribute the 3 topics to those teams.

Let's think of the 3 topics as 3 slots to fill. If we have 3 groups for 3 slots, there are 3! different ways to arrange them.

So, we have 90 possible groups and 6 different ways to assign them to topics, giving us:

90 * 6 = 540 possibilities.

Nice catch, Stuart
Caught me sleepting

Brent Hanneson - Creator of GMATPrepNow.com

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Wed Jan 13, 2010 11:00 am

Thanks Stuart and Brent for your inputs :)

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by Brent@GMATPrepNow » Wed Jan 13, 2010 11:39 am

Hold the phone

After taking a second look at the question, I believe that the answer is, indeed, 90 (C).

There are 15 ways to divide 6 people (ABCDEF) into 3 teams.
They are:
AB CD EF
AB CE DF
AB CF DE
AC BD EF
AC BE DF
AC BF ED
.
.
.
I'll stop here. We can see that, once we have paired A with someone, the remaining 4 students can be paired in 3 ways.
Since A can have 5 different partners (and, for each of these pairings, we have 3 subesquent pairings possible), there are a total of 15 ways to divide 6 people into 3 teams.

Once we have divided the students into 3 teams, we can see that there are 3! ways to assign each team a different topic

So, we get 15 x 3! = 90

My original method divided the 6 students into 3 teams AND assigned topics at the same time.

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by Brent@GMATPrepNow » Wed Jan 13, 2010 5:57 pm

What's the source of this question?

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Thu Jan 14, 2010 7:25 am

well brent..a friend from my math tuitions shared this with me !!

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apoorva.srivastva: Master | Next Rank: 500 Posts; Posts: 146; Joined: Wed Aug 27, 2008 5:41 am; Thanked: 3 times

by apoorva.srivastva » Sat Apr 24, 2010 7:23 am

i just tried to solve it all over again and realised that E is the answer..please correct me if i am wrong.

ways to divide a team of 6 students into 3 groups of 2 each = 6!/ (2!)^3 = 90

and now they can be assigned to 3 different topics in 3! ways.

Therefore the total number of arrangements = 90 * 3! = 540

So ans is 540 (E)

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sy323: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Fri Apr 20, 2012 11:28 am; Thanked: 2 times

by sy323 » Sun Jun 30, 2013 10:18 am

I'm actually having trouble understanding the last step. which is 3 topics assigned to the 3 groups why is this 3!? what is the formula is this case?

I am unable to understand, i'm sure this is a relatively simple concept

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pmFollowed by: 4 members; Thanked: 29 times

by vipulgoyal » Sun Jun 30, 2013 10:06 pm

I believe Brent is right

6 people can be arranged in 3 groups of 2-2 each by 15 ways and then projects can be allocated by 3! ways -- A,B,C,D,E,F; 5+4+3+2+1 = (15 ways * 3! = 90)
ab,ac,ad,ae,af,bc,bd,,be......ef ( As Brent explained ab and ba are same)

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by Brent@GMATPrepNow » Mon Jul 01, 2013 9:27 pm

apoorva.srivastva wrote:Six students are equally divided into 3 groups, then, the three groups were assigned to three different
topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

Let's say we have 3 topics: Topic A, Topic B and Topic C.
Our goal is to assign 2 people to each topic.
Let's take this task and break it into stages.

Stage 1: Select 2 people for Topic A.
Since the order in which we select these people does not matter (e.g., selecting Joe then Al is the same as selecting Al then Joe), we can use combinations.
There are 6 people, and we must select 2. This can be accomplished in 6C2 ways (15 ways).

Stage 2: Select 2 people for Topic B.
Once again, order does not matter.
There are 4 people remaining, and we must select 2. This can be accomplished in 4C2 ways (6 ways).

Stage 3: Select 2 people for Topic C.
There are only 2 people remaining, so this can be accomplished in 1 way

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus assign 2 people to each topic) in (6)(5)(1) ways ([spoiler]= 90 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Brent Hanneson - Creator of GMATPrepNow.com

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vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pmFollowed by: 4 members; Thanked: 29 times

by vipulgoyal » Tue Jul 02, 2013 1:43 am

hi Brent, In one more source I found the same qustion with 540 ans
6c2*4c2*3p3=540
does this require second thought??

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Jul 02, 2013 5:22 am

vipulgoyal wrote:hi Brent, In one more source I found the same qustion with 540 ans
6c2*4c2*3p3=540
does this require second thought??

Here's my interpretation of the question:
We have 3 topics: Topic A, Topic B and Topic C. Our goal is to assign 2 people to each topic.
If this interpretation is correct, then I believe the answer is 90.

Let's first divide the 6 people into to three groups of 2. We'll list all 15 possibilities

1) AB CD EF
2) AB CE DF
3) AB CF DE
4) AC BD EF
5) AC BE DF
6) AC BF ED
7) AD BC EF
8) AD BE CF
9) AD BF CE
10) AE BC DF
11) AE BD CF
12) AE BF CD
13) AF BC DE
14) AF BD CE
15) AF BE CD

For each of the 15 scenarios, we can choose a pair to cover Topic A in 3 ways (since there are 3 pairs to choose from). After that, we can choose a pair to cover Topic B in 2 ways (since there are 2 pairs remaining). And we can choose a pair to cover Topic C in 1 way.

So, we can assign the 3 Topics in (15)(3)(2) (1) ways (= 90 ways)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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