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

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

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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

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Last edited by Bunuel on 28 Feb 2012, 13:07, edited 2 times in total.
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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.
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Re: arrangements made [#permalink]

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New post 17 Feb 2011, 05:29
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.
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New post 28 Feb 2012, 12:57
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
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combination [#permalink]

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New post 05 Apr 2012, 11:51
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
please explain your reasoning


thanks

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Re: combination [#permalink]

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New post 05 Apr 2012, 12:09
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
please explain your reasoning


thanks

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Re: combination [#permalink]

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New post 05 Apr 2012, 12:33
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
please explain your reasoning


thanks

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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
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Re: combination [#permalink]

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New post 05 Apr 2012, 15:57
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
please explain your reasoning


thanks

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Merging similar topics. Please ask if anything remains unclear.

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

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New post 07 Jan 2013, 10:51
ededed OR dedede
3*3*2*2*1*1 + 3*3*2*2*1*1
= 72
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Re: 3 dwarves and 3 Elves sit down in a row of 6 chairs. If no [#permalink]

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New post 07 Jan 2013, 10:56
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
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Three dwarves and three elves sit down in a row of six chairs. I [#permalink]

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New post 22 Jan 2013, 02:34
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)?
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Re: Three dwarves and three elves sit down in a row of six chair [#permalink]

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New post 22 Jan 2013, 20:17
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/)
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Permutation & Combination question [#permalink]

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New post 15 Dec 2013, 08:27
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.

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Re: Permutation & Combination question [#permalink]

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New post 15 Dec 2013, 18:49
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



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Re: Permutation & Combination question [#permalink]

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New post 16 Dec 2013, 00:13
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


Merging similar topics. Please ask if anything remains unclear.
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New post 12 Aug 2014, 11:44
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


Merging similar topics. Please ask if anything remains unclear.


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).
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