Q from MGMAT guide 4, page 76, Q14:

Three gnomes and three elves sit down in a row of six 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?

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I tried to approach this problem in a similar way taught earlier in the chapter: calculate all of the options without restrictions, and then delete out the restrictions. This gave me a sum much higher than the actual answer. I did not understand the book's explanation and would like to know if/how to solve this using the "work backwards" approach.

First please put the Guide Name and edition as mentioned in the Forum Readme in the question. In this case it was the WT Guide 4th edition

This is a basic counting problem.


Say the 6 gnomes and elves are standing in the corner and you have to seat them

For the first chair you have 6 choices ( 3 gnomes and 3 elves)

For the 2nd chair you have 3 choices (if the first one you chose was a gnome then the only choice is one of the 3 elves. If the first one you chose was an elf then the only choice is one of the 3 gnomes)

For the 3rd chair you have 2 choices (At this point one each of a gnome and elf are seated. That means 2 gnomes and 2 elves are left. You can choose from either 2 gnomes or 2 elves depending on how you started - but in either case your choice is only 2)

For the 4th chair you have 2 choices. (At this point you have two scenarios GEG or EGE in the chairs. Either way you have a choice between two elves in the first case or two gnomes in the second case. your choice is still 2)

For the 5th chair only 1 choice [ simply choose either a gnome or elf depending on who can sit)

For the 6th chair the last choice - still 1

to see how many ways you can seat them simply multiply your choices

6 * 3 * 3 * 2 * 2 * 1 *1 = 72

"Work backwards" will entail listing 72 combinations which you really don't want.

Please note that I haven't read what the "SLOT method" is so it could be different.

Makes sense, thanks. FYI, there is a typo in your final formula, you have an extra "3".

Let's try a more intuitive approach.

Let's call the gnomes A, B, and C. Let's call the elves X, Y, and Z.

If we want to make sure gnomes and elves don't sit with the same kind, then there are two arrangements:
(1) GEGEGE
or
(2) EGEGEG

Let's just take a look arrangement (1). If we take a look ONLY at the gnomes, there are six ways (or 3!) they can be seated from left to right:
A-B-C-
A-C-B-
B-A-C-
B-C-A-
C-A-B-
C-B-A-

if we take a look at ONLY the elves, there are also six ways they could be seated from left to right.

If there are 6 ways the gnomes could be seated and 6 ways for the elves to be seated, then there are 6 * 6 = 36 options for arrangement (1).

Arrangement (2) is like arrangement (1) except we start with elves first. Using the same reasons, there are 36 options for arrangement (2).

The total ways the elves and gnomes can be arranged are 36+ 36 = 72.

_________________
Ben Ku
Instructor
ManhattanGMAT

The last section of Chapter 4 discusses "Arrangements with Constraints" (Guide 4, 4th Ed, Page 73) and introduces the "Glue Method”.

In a specific example on the page listed, two of the people cannot sit together at a movie. These two people are "stuck together" during the explanation (i.e. - instead of 6 people, 5 people are used for the 2nd part of the calculation).

Can the Glue Method be used on the "Three Gnomes and Three Elves" problem?
Thanks!

No, the "Glue Method" cannot be used because there are no constraints in the problem where two gnomes or elves MUST sit next to each other. (We are only given constraints about who must not sit next to each other.)

_________________
Ben Ku
Instructor
ManhattanGMAT

OK Ben, thanks!

you're welcome!

_________________
Ben Ku
Instructor
ManhattanGMAT

Ben,

Can you explain in detail where glue method should be used and where not.As movie question mentioned by one user here also has same conditions" should not sit together"

Ben,

Can you explain in detail where glue method should be used and where not.As movie question mentioned by one user here also has same conditions" should not sit together"

Ron,

Can you at least help here?

The "glue method" is generally used when there is some sort of constraint in which two people don't want to sit next to each other, or two things can't be placed next to each other. In these cases we are looking for the number of acceptable outcomes. The easiest way to find them is to calculate ALL possible outcomes, then "glue" the people who refuse to sit next to each other together, and find the number of unacceptable possibilities in which those people are in fact next to each other. Then take subtract unacceptable from total.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Why is it, when you use the slot method, it is not necessary to multiply by 2? For example, if you figure out the number of possibilities for GEGEGE, why do you not have to do it again for EGEGEG?

Also, does the glue method not work here because it involves groups of 3 rather than 2 people that do not want to sit together?

_________________
Jamie Nelson
ManhattanGMAT Instructor