Three dwarves and three elves sit down in a row of six chairs. If no

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?

A. 18
B. 36
C. 48
D. 72
E. 96


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.

Answer: D.
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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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/10 ... inatorics/)
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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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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

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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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 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.

Does the formula 6!/3!3! only work if they are sitting next to each other?

e.g. sitting DDDEEE or EEEDDD.

Or is it due to having 2 different sub-sets that we can't use such formula.

Thank you for the help.

6!/(3!3!)=20 is the number of arrangements of 6 letters EEEDDD, where 3 D's and 3 E's are identical:
EEEDDD;
EEDEDD;
EDEEDD;
DEEEDD;
...
DDDEEE.

But in our original question we don't have 3 identical D's and 3 identical E's. Also, we need only those arrangements where no dwarf will sit next to another dwarf and no elf will sit next to another elf.

In case of further questions please post here: three-dwarves-and-three-elves-sit-down-in-a-row-of-six-109445.html
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Case 1: G E G E G E - Total ways = 3!*3! = 36

Case 2: E G E G E G - Total ways = 3!*3! = 36

Total Ways of arrangements = 36+36 = 72

Answer: Option D

Basically the above objective would be achieved if elves and genomes sit alternately.
Let the chair numbers be 1, 2, 3, ...6.
So we can make them sit in two ways:

Case 1. Elves on chair numbers 1, 3, 5. Genomes on chair numbers 2, 4, 6
This can be done in 3! * 3! = 36 ways

Case 2. Genomes on chair numbers 1, 3, 5. Elves on chair numbers 2, 4, 6
This can be done in 3! * 3! = 36 ways

Thus total ways = 36+36 = 72

Hence answer is D

Bunuel Hi, I have one doubt please help me.
I don't know why this approach is wrong.
we have four places where we can put elf *D*D*D* 4C2*3!*3!?
Please help me why this is wrong?

I guess you meant 4C3*3!*3!.

The point is that dwarves and elves must sit either DEDEDE or EDEDED but if you choose 3 * out of 4 you can get *DD*D*, which violates the restriction we have that no dwarf will sit next to another dwarf and no elf will sit next to another elf. So, your way gives more cases, which are not allowed.

Hope it's clear.
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We can see that one seating arrangement of gnomes (G) and elves (E) can be:

GEGEGE

The first G and E each has 3 choices; the second G and E each has 2 choices and the last G and E each has 1 choice. Thus the number of ways to have the “GEGEGE” seating arrangement is:

3 x 3 x 2 x 2 x 1 x 1 = 36

However, the seating arrangement can also be EGEGEG, and there will also be 36 such arrangements. Thus, the total number of seating arrangements is 36 + 36 = 72.

Answer: D
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One quick question,

Why we have not multiplied ?

I , initially multiplied but later on checked that the answer is not matching so I added !

How to quickly identify in these questions whether we need to add or multiply when nothing as such is mentioned

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The seating arrangement will be either G-E-G-E-G-E or E-G-E-G-E-G, it can't be both at the same time. Whenever you have OR this means addition. Whenever you have AND this means multiplication.

gmatophobia

help me with this one?
what if find the dwarf will sit next to another dwarf and elf will sit next to another elf and substract it from total possibility?
i didnt get the correct answer. is it possible to do it by this method?

AnujL

You can do that, but that will be way too many cases to subtract.

Lets see what could be the possible cases be -
1) Two dwarves together
2) Three dwarves together
3) Two Elves together
4) Three elves together

Also as there is likely overlap between each case that needs to factored in as well.

A better and a simpler method will be to use the filling space approach.

Assumed that the chairs are arranged as shown below -

+ _ + _ + _ +

We can select either the dwarves or Elves first and place them either in the + chairs or the _ chairs. This can be done in 2 ways.

Next we can arrange each of them in 3! ways.

So total ways = 2 * 3! * 3! = 72 ways
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