6 children are sitting on a merry-go-round, in how many ways can you switch seats so that no one sits opposite the person who is opposite to them now?

Question

I'm preparing for an exam and I would appreciate your help if you can tell me if there's a mistake in my solution.

Question

$6$ children are sitting on a merry-go-round:

Now, we need to change their seats s.t kid $1$ doesn't sit opposite kid $4$, kid $2$ doesn't sit opposite kid $5$ and kid $3$ doesn't sit opposite kid $6$.

$A_1 $ is the set of all their ways to sit in a circle s.t kid $1$ sits opposite kid $4$.
$A_2 $ is the set of all their ways to sit in a circle s.t kid $2$ sits opposite kid $5$.
$A_3 $ is the set of all their ways to sit in a circle s.t kid $3$ sits opposite kid $6$.

$U$ is the set of all their ways to sit in a circle.

We need to count $U\setminus (A_1 \cup A_2 \cup A_3)$.

$$U\setminus (A_1 \cup A_2 \cup A_3)= 5! - 3 \cdot 4! -3 \binom{3}{2}\cdot 2\cdot 2-2^3=5!-28 = 92$$

Explanation:

• The $6$ children are sitting on a merry-go-round.
  • That defines an order on them.
• I use symmetric inclusion and exclusion to solve the problem.

I get confused while trying to calculate $A_1$ for exapmle (I think the first way is correct)

Answer 1

Your question has been answered properly already (explaining nicely PIE, i.e. the principle of inclusion/exclusion), but let me offer you an alternative solution.

First place kid1.

Then place kid4 on a position not opposed to the position of kid1 ($4$ possibilities).

After that exactly one pair of open seats opposed to eachother exists.

Let one of these two seats be occupied by one of the $4$ remaining kids ($4$ possibilities).

Let the other of them be occupied by a kid that was not sitting originally opposed to the kid we placed in the former move ($2$ possibilities).

Now place the remaining two kids ($2$ possibilities).

So there are:$$4\times4\times2\times2=64$$

This agrees with the answer of N.F. Taussig.

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You could say that this solution is more direct and it is always nice to have more than one approach. Especially because then you can check yourself.

Nevertheless I would recommend the solution that is based on PIE because it is more constructive.

Answer 2

Since you are using the Inclusion-Exclusion Principle, the signs of the terms should alternate.

Relative to kid $1$, there are $5!$ ways to seat the remaining people.

A pair of kids who were originally sitting opposite to each other are sitting opposite to each other again: There are three ways to select the pair. Use the younger member of that pair as our reference point. There is one way to seat the other member of that pair opposite to that kid. There are $4!$ ways to seat the remaining kids as we proceed clockwise around the merry-go-round from the kid we are using as our reference point. Hence, there are $\binom{3}{1}4!$ such seating arrangements.

If we subtract this amount from the total, we will have subtracted too much since we will have subtracted each seating arrangement in which two pairs of kids who were originally sitting opposite to each other are still sitting opposite to each other in the new arrangement twice, once for each way of designating one of those pairs as the pair of kids who are still sitting opposite to each other. Therefore, we need to add the number of arrangements in which two pairs of kids who were originally sitting opposite to each other are opposite to each other.

Two pairs of kids who were originally sitting opposite to each other are sitting opposite to each other again: There are $\binom{3}{2}$ ways to select which two pairs of kids who were originally sitting opposite to each other are also sitting opposite to each other in the new arrangement. Observe that this also forces the third pair of kids to sit opposite to each other. Take the youngest child as our reference point. The person who was originally sitting opposite to that child must sit opposite to that child. There are four ways to select the person who sits to the immediate left of the youngest child. Seating that person also determines where the other member of that pair sits. That leaves two people. There are two ways to select the person who sits to the immediate right of the youngest child. Seating that person also determines where the remaining person sits. Hence, there are $\binom{3}{2} \cdot 4 \cdot 2$ such seating arrangements.

However, if we add that amount to our running total, we will have added too much. We have not subtracted the number of seating arrangements in which all three pairs of people who were originally opposite to each other are again opposite to each other since we first subtracted such arrangements three times, once for each of the three ways we could have designated one of those three pairs as the pair of people who sit opposite to each other in the new seating arrangement, and then added them three times, once for each of the $\binom{3}{2}$ ways of designating two of those three pairs as the two pairs of kids who are again sitting opposite to each other in the new arrangement. Therefore, we must subtract the number of arrangements in which all three pairs of kids who were originally sitting opposite to each other are still sitting opposite to each other.

Three pairs of kids who were originally sitting opposite to each other are sitting opposite to each other again: The argument given above shows that there are $4 \cdot 2$ such seating arrangements.

Hence, by the Inclusion-Exclusion Principle, the number of seating arrangements in which no two kids who were originally sitting opposite to each other are still sitting opposite to each other is $$5! - \binom{3}{1}4! + \binom{3}{2} \cdot 4 \cdot 2 - 4 \cdot 2$$

Answer 3

To simplify presentation, I shall label the kids as $ABC$, and the initial arrangement as $\dfrac{ABC}{ABC}$
meaning that $A,B,C$ are diametrically opposite to $A,B,C$ respectively.

With the first $A$ fixed, possibilities for the first half of circle are

$AAB\;\;AAC\;\;ABA\;\;ABB\;\;ABC\;\;ACA\;\;ACB\;\;ACC$

For the second half, $ABC$ is known to have $2$ derangements, and so do each of the other blocks (since there are two possible places for the letter that occurs only once in the first half)

thus there are $16$ valid blocks, and since the first $A$ can't be switched, but the $B's $ and $C's$ can,

answer $= 16\times 2^2 = 64$