Dwarves and Elves PERMUTATIONS (MGMAT BOOK QUESTION)

[Stockmoose16]

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 will sit next to another elf, in how many different ways can the elves and dwarves sit?


I've tried two methods and get two different answers. Someone please explain what I'm doing wrong in method #2. Both methods should give the same answer.

METHOD #1:

Seat-by-Seat method:

6 (# of options available, we'll start w/dwarf) * 3 (# of remaining elves) *2 (# of remaining dwarves) *2 (# of remaining elves) *1 (# of remaining dwarves) * 1 (# of remaining elves) = 72

METHOD #@

Enumeration

The elves can be arranged in any of the following manner:

E1D1, D1E1
E1D2, D1E2
E1D3, D1E3
E2D1, D2E1
E2D2, D2E2
E2D3, D2E3
E3D1, D3E1
E3D2, D3E2
E3D3, D3E3

So, we have 18 unique ways the Elves and Dwarves can be paired. In addition there are 3 positions that each of these paired sets can sit in (i.e seats 1&2, 3&4, 5&6). Thus, the answer should be 18 pairs * 3 different arrangements for each pair. That gives me 54. Why am I coming up with the wrong answer using this method?

[tendays2go]

the first gives you combinations while second gives you permutations.
since a restriction condition is specified here, it is a permutation problem:

consider sitting order of: dedede
then, 3.3.2.2.1.1 = 36 ways
PS: both dwarfs and elves seat independently of each other
now, it could be ededed as well
i.e. another 3.3.2.2.1.1 = 36 ways

hence total are 72 ways

[4meonly]

DEDEDE
D can permutate in 3! ways, E can permutate in 3! ways. So we have 3!*3!

we also can have
EDEDED

So we need to multiply 3!*3! by 2

we have 3!*3!*2 = 72

[Stockmoose16]

the first gives you combinations while second gives you permutations.
since a restriction condition is specified here, it is a permutation problem:

consider sitting order of: dedede
then, 3.3.2.2.1.1 = 36 ways
PS: both dwarfs and elves seat independently of each other
now, it could be ededed as well
i.e. another 3.3.2.2.1.1 = 36 ways

hence total are 72 ways

If it's a permutations problem, then why does method #1, which uses combinations, come out with the correct answer, but method #2, which uses enumeration (permutations) doesn't come out correctly. What is wrong with method #2?

[mrinal2100]

cant we have this situation XyXyXyX.....there r four X position where Dwarf can fix......

as a result it would be 4P3 and 3 poitions will be left ie..3!... total no of permutation =4p3*3!

[kwstewa]

If anyone has a minute, I'd like to enlist the people who have posted here to help with the following problem:

https://www.beatthegmat.com/mgmat-proble ... html#82919

Someone else posted this problem, but did not post the answer. Here it is:

Six mobsters have arrived at the theater for the premiere of the film “Goodbuddies.” One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

To me, this seems like a standard permutation problem, but the twist is that because Frankie insists on standing behind Joey, Joey cannot stand behind Frankie in line. This means that only five of the mobsters potentially could stand in the last position, 6 mobsters potentially could stand in positions 2 through 5, and only 5 mobsters potentially could stand in the front (because Frankie must stand behind Joey). Does anyone know how you account for this restriction? If any of the mobsters could stand in any position in line, it seems you would solve by multiplying the following:

6*5*4*3*2*1 = 720 permutations

But I don't know how to account for the fact that Frankie must stand behind Joey.

[mental]

i have just posted a reply on mobsters chain
my ans 360
(6P2)*4!/2

[sandeepraghuvanshi]

3d,3e,2d,2e,1d,1e=36ways
or
3d.............


total=72 ways

[Matt@VeritasPrep]

I think we didn't need to revive this thread