Word Translation :Combinatorics - Chapter 4, Question 14

[aksharma]

The question is - 3 dwarves and 3 elves sit down in a row of 6 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 understand the solution to the problem. Just wondering if it would be possible to solve it using an alternate method - where you calculate the different ways all 6 can sit and the subtract the ways in which the dwarves and elves are sitting next to each other? (similar to the explanation on Page 59)

Thanks,
Akshay.

[shaji]

The question is - 3 dwarves and 3 elves sit down in a row of 6 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 understand the solution to the problem. Just wondering if it would be possible to solve it using an alternate method - where you calculate the different ways all 6 can sit and the subtract the ways in which the dwarves and elves are sitting next to each other? (similar to the explanation on Page 59)

Thanks,
Akshay.

Yes!!! U can. Try it out and let know of any difficulty.

[shaji]

The question is - 3 dwarves and 3 elves sit down in a row of 6 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 understand the solution to the problem. Just wondering if it would be possible to solve it using an alternate method - where you calculate the different ways all 6 can sit and the subtract the ways in which the dwarves and elves are sitting next to each other? (similar to the explanation on Page 59)

Thanks,
Akshay.

Yes!!! U can. Try it out and let know of any difficulty.

In case of any difficulty,pleaspost all answer choices so that the best approach can be suggested. The direct approach is what I would prefer in the event that the correct answer choices are not vailable or more precisely thecorrect answer is not staring at you!!!.
The correct answer is indeed 144.

[RonPurewal]

I understand the solution to the problem. Just wondering if it would be possible to solve it using an alternate method - where you calculate the different ways all 6 can sit and the subtract the ways in which the dwarves and elves are sitting next to each other? (similar to the explanation on Page 59)

Thanks,
Akshay.

hey -

yes, you could solve the problem by that sort of approach, but it would be gruesomely ugly - you'd have to eliminate the vast majority of the possibilities - 90% of them, to be exact.

what's more, you'd have to come to the same realization necessary for the book's solution (namely, that the arrangement has to be either e-d-e-d-e-d or d-e-d-e-d-e) but, instead of arriving at that conclusion directly, you'd have to arrive at it by eliminating all 18 of the other possibilities (dddeee, ddedee, ..., eeeddd).

horrible.

if you want an alternate solution, you could do this:
- first realize that the solution has to be either d-e-d-e-d-e or e-d-e-d-e-d
- for each of these two scenarios, note that there are (3!)(3!) = 36 ways of arranging the elves & dwarves (3! for the elves, 3! for the dwarves, and multiply because they're independent)
- so it's 36 + 36, or 72

[shaji]

I understand the solution to the problem. Just wondering if it would be possible to solve it using an alternate method - where you calculate the different ways all 6 can sit and the subtract the ways in which the dwarves and elves are sitting next to each other? (similar to the explanation on Page 59)

Thanks,
Akshay.

hey -

yes, you could solve the problem by that sort of approach, but it would be gruesomely ugly - you'd have to eliminate the vast majority of the possibilities - 90% of them, to be exact.

what's more, you'd have to come to the same realization necessary for the book's solution (namely, that the arrangement has to be either e-d-e-d-e-d or d-e-d-e-d-e) but, instead of arriving at that conclusion directly, you'd have to arrive at it by eliminating all 18 of the other possibilities (dddeee, ddedee, ..., eeeddd).

horrible.

if you want an alternate solution, you could do this:
- first realize that the solution has to be either d-e-d-e-d-e or e-d-e-d-e-d
- for each of these two scenarios, note that there are (3!)(3!) = 36 ways of arranging the elves & dwarves (3! for the elves, 3! for the dwarves, and multiply because they're independent)
- so it's 36 + 36, or 72

72 is indeed correct!!!. 144 includes two d's being together.