It is currently 19 Oct 2017, 13:37

### 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

# Three dwarves and three elves sit down in a row of six

Author Message
TAGS:

### Hide Tags

Manager
Joined: 10 Nov 2010
Posts: 247

Kudos [?]: 401 [0], given: 22

Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
Three dwarves and three elves sit down in a row of six [#permalink]

### Show Tags

16 Feb 2011, 13:13
4
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:19) correct 0% (00:00) wrong based on 3 sessions

### HideShow timer Statistics

Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to another dwarf and no elf wil sit next to another elf, in how many different ways can the elves and dwarves sit?

as per my solution

on first chair from 3dwarves any one can sit on another chair from 3elves any one can sit nd so on

3*3*2*2*1*1=36

But OA is
[Reveal] Spoiler:
72

_________________

The proof of understanding is the ability to explain it.

Last edited by Bunuel on 28 Feb 2012, 14:07, edited 2 times in total.
Edited the question

Kudos [?]: 401 [0], given: 22

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [1], given: 12183

### Show Tags

16 Feb 2011, 13:27
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to another dwarf and no elf wil sit next to another elf, in how many different ways can the elves and dwarves sit?

In order to meet the restriction dwarves and elves must sit either DEDEDE or EDEDED. There are 3!*3!=36 arrangements possible for each case (3! arrangements of dwarves and 3! arrangements of elves), so total ways to sit are 2*36=72.
_________________

Kudos [?]: 128901 [1], given: 12183

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17361 [1], given: 232

Location: Pune, India
Re: Three dwarves and three elves sit down in a row of six chair [#permalink]

### Show Tags

22 Jan 2013, 21:17
1
KUDOS
Expert's post
manimgoindowndown wrote:
How would we do this using COMBINATIONS/C as opposed to permutations or just counting out possibilities for each of the 6 slots (ie 6x3x2x2)?

You need to arrange the 6 people in 6 places. You can think in terms of combinations in the following way:

For the first chair, we select any one person of the 6 in 6C1 ways.
For the second chair, we select any one from the 3 in 3C1 ways (e.g. if an elf is sitting on the first chair, we need to choose out of the 3 dwarves only)
For the third chair, we select one of the leftover 2 (e.g. one elf is already sitting in first chair) in 2C1 ways
For the fourth chair, we select one of the leftover 2 (e.g. one dwarf is already sitting in the second chair and we need to choose of the remaining 2) in 2C1 ways
For the fifth and sixth chairs, there is only one choice each.

Answer: 6C1 * 3C1 * 2C1 * 2C1 = 6*3*2*2 = 72
(note that this is the basic counting principle itself. For more, check: http://www.veritasprep.com/blog/2011/10 ... inatorics/)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17361 [1], given: 232

Current Student
Status: Up again.
Joined: 31 Oct 2010
Posts: 527

Kudos [?]: 530 [0], given: 75

Concentration: Strategy, Operations
GMAT 1: 710 Q48 V40
GMAT 2: 740 Q49 V42

### Show Tags

17 Feb 2011, 06:29
GMATD11 wrote:
Pls chk the image

as per my solution

on first chair from 3dwarves any one can sit on another chair from 3elves any one can sit nd so on

3*3*2*2*1*1=36

But OA is 72

You forgot to multiply the answer with 2. You see, either one the dwarves can take the first seat or one of the elves can take it. If a dwarf occupies the first seat, the other dwarves will sit on the 3rd and 5th seat. Elves will sit on 2nd, 4th and 6th seats. So as you have rightly calculated above, there will be 36 seating arrangements.

In the event where an elf takes the first seat, there will be as many combinations possible albeit the seats will change, with the elves taking the odd numbered seats and the dwarves taking the even numbered ones.

So total number of combinations= 36+36= 72.
_________________

My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html

Kudos [?]: 530 [0], given: 75

Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 261

Kudos [?]: 1226 [0], given: 48

### Show Tags

28 Feb 2012, 13:57
there are only two patterns: dedede, ededed. each will return same number of combinations.

combinations = 3*3*2*2*1*1*(2) = 72 ... multiply (2) because there are two patterns
_________________

press +1 Kudos to appreciate posts

Kudos [?]: 1226 [0], given: 48

Intern
Joined: 05 Apr 2012
Posts: 46

Kudos [?]: 37 [0], given: 12

### Show Tags

05 Apr 2012, 12:51
hello

IF 3 DWARFS AND THREE ELVES SIT DOWN IN A ROW OF 6 chairs . IF NO DWARFS WILL SIT NEXT TO ANOTHER DWARFS AND NO ELFS WILL SEAT NEXT TO ANOTHER ELFS
IN HOW MANY WAYS CAN THE ELVES AND DWARFS SIT?

answer to this practice exercises is 72

thanks

best regards

Kudos [?]: 37 [0], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

### Show Tags

05 Apr 2012, 13:09
keiraria wrote:
hello

IF 3 DWARFS AND THREE ELVES SIT DOWN IN A ROW OF 6 chairs . IF NO DWARFS WILL SIT NEXT TO ANOTHER DWARFS AND NO ELFS WILL SEAT NEXT TO ANOTHER ELFS
IN HOW MANY WAYS CAN THE ELVES AND DWARFS SIT?

answer to this practice exercises is 72

thanks

best regards

P.S. Please turn Caps Lock off when posting.
_________________

Kudos [?]: 128901 [0], given: 12183

Intern
Joined: 05 Apr 2012
Posts: 46

Kudos [?]: 37 [0], given: 12

### Show Tags

05 Apr 2012, 13:33
Bunuel wrote:
keiraria wrote:
hello

IF 3 DWARFS AND THREE ELVES SIT DOWN IN A ROW OF 6 chairs . IF NO DWARFS WILL SIT NEXT TO ANOTHER DWARFS AND NO ELFS WILL SEAT NEXT TO ANOTHER ELFS
IN HOW MANY WAYS CAN THE ELVES AND DWARFS SIT?

answer to this practice exercises is 72

thanks

best regards

P.S. Please turn Caps Lock off when posting.

Thanks my approach was the total number of way the 6 could sit minus the number of way the dwarf and the elfs can sit together
6!-3!x3!

i dont understand my mistake
best regards

Kudos [?]: 37 [0], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

### Show Tags

05 Apr 2012, 16:57
keiraria wrote:
Bunuel wrote:
keiraria wrote:
hello

IF 3 DWARFS AND THREE ELVES SIT DOWN IN A ROW OF 6 chairs . IF NO DWARFS WILL SIT NEXT TO ANOTHER DWARFS AND NO ELFS WILL SEAT NEXT TO ANOTHER ELFS
IN HOW MANY WAYS CAN THE ELVES AND DWARFS SIT?

answer to this practice exercises is 72

thanks

best regards

P.S. Please turn Caps Lock off when posting.

Thanks my approach was the total number of way the 6 could sit minus the number of way the dwarf and the elfs can sit together
6!-3!x3!

i dont understand my mistake
best regards

The problem with that solution is that 3!*3! does not give you all cases when dwarves and elves sit together. 3!*3! counts only the cases when all 3 dwarves and all 3 elves are sitting together for example all cases for DDDEEE. But since we are asked to find the ways to arrange them so that no dwarf will sit next to another dwarf and no elf wil sit next to another elf then there are many other cases possible when this condition is violated: EEEDDD, EEDDED, DDEEDE, ...
_________________

Kudos [?]: 128901 [0], given: 12183

Manager
Joined: 13 Oct 2012
Posts: 69

Kudos [?]: -9 [0], given: 0

Schools: IE '15 (A)
GMAT 1: 760 Q49 V46
Three dwarves and three elves sit down in a row of six chairs. I [#permalink]

### Show Tags

07 Jan 2013, 11:51
ededed OR dedede
3*3*2*2*1*1 + 3*3*2*2*1*1
= 72

Kudos [?]: -9 [0], given: 0

Manager
Joined: 18 Oct 2011
Posts: 90

Kudos [?]: 91 [0], given: 0

Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Re: 3 dwarves and 3 Elves sit down in a row of 6 chairs. If no [#permalink]

### Show Tags

07 Jan 2013, 11:56
you can either have the following arragements of seating positions
EDEDED or DEDEDE

Therefore for the first arragement you would have 3!x3! ways to sit the 6 people. Since there are two possible seating arragememts you would have 2(3!x3!) = 72

Kudos [?]: 91 [0], given: 0

Manager
Joined: 07 Feb 2011
Posts: 106

Kudos [?]: 63 [0], given: 45

Three dwarves and three elves sit down in a row of six chairs. I [#permalink]

### Show Tags

22 Jan 2013, 03:34
How would we do this using COMBINATIONS/C as opposed to permutations or just counting out possibilities for each of the 6 slots (ie 6x3x2x2)?
_________________

Kudos [?]: 63 [0], given: 45

Intern
Joined: 29 Aug 2013
Posts: 14

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

### Show Tags

15 Dec 2013, 09:27
Hello forum,

how can you solve this question using either permutation or combination?

3 gnomes and 3 elves sit down in a row of 6 chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

This is a question from one of the MGMAT books and their solutions does not mention a way to solve this by using either the formula for combination or permutation. I could not solve it on my own, so I would kindly ask you to help me with this.

Best

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

SVP
Joined: 14 Apr 2009
Posts: 2138

Kudos [?]: 1601 [0], given: 8

Location: New York, NY
Re: Permutation & Combination question [#permalink]

### Show Tags

15 Dec 2013, 19:49
You might break this down. At the highest level there's only 2 sets:

Seats #1, 2, 3, 4, 5, 6 would be:
Gnome, Elf, Gnome, Elf, Gnome, Elf

Since gnomes can't sit next to gnomes and elves can't sit next to elves, the only other alternative for seats would be:
Elf, Gnome, Elf, Gnome, Elf, Gnome

We did that by moving the gnome to seat #2. But moving it to seat #3 would be a repeat of the first set mentioned. So there's only 2.

Now we have 2 * (TBD)

For each of the 2 sets above, the gnomes can be arranged and the elves can be fixed in their position.
So just line gnome1, gnome2, gnome3 and then rearrange them.

That's 3P3 = 3! = 3*2 = 6
So if the elves are fixed, there are 6 gnome variations.

But wait...the elves can vary too:
They also can be 3P3 = 6 with the gnomes fixed in their position.

So do 6 * 6 = 36 variations

But wait...we had 2 arrangements in the very beginning. One where gnome was at the very beginning and another where gnome was in the #2 spot.

SO take 36 * 2= 72

72 arrangements: 2 sets of 6 variations on 6 variations

For more information on permutations/combinations, try this free lesson on permutations/combinations from GMAT Pill.

Kudos [?]: 1601 [0], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128901 [0], given: 12183

Re: Permutation & Combination question [#permalink]

### Show Tags

16 Dec 2013, 01:13
mott wrote:
Hello forum,

how can you solve this question using either permutation or combination?

3 gnomes and 3 elves sit down in a row of 6 chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

This is a question from one of the MGMAT books and their solutions does not mention a way to solve this by using either the formula for combination or permutation. I could not solve it on my own, so I would kindly ask you to help me with this.

Best

_________________

Kudos [?]: 128901 [0], given: 12183

Intern
Joined: 12 Aug 2014
Posts: 2

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

Three dwarves and three elves sit down in a row of six [#permalink]

### Show Tags

12 Aug 2014, 12:44
Bunuel wrote:
mott wrote:
Hello forum,

how can you solve this question using either permutation or combination?

3 gnomes and 3 elves sit down in a row of 6 chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

This is a question from one of the MGMAT books and their solutions does not mention a way to solve this by using either the formula for combination or permutation. I could not solve it on my own, so I would kindly ask you to help me with this.

Best

I'm trying to solve this in manner similar to your previous post ("Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements"... I am unable to post the link). How would you apply the rule where you find the total # of arrangements and then subtract the # of violations?

Can you please explain the difference? I cannot seem to solve this question (dwarves & elves) like the previous one (movie seats).

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

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16651

Kudos [?]: 273 [0], given: 0

Re: Three dwarves and three elves sit down in a row of six [#permalink]

### Show Tags

16 Aug 2016, 10:36
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: Three dwarves and three elves sit down in a row of six   [#permalink] 16 Aug 2016, 10:36
Display posts from previous: Sort by