Let m be the number of pairs, so that $n=2m$.
Let S be the set of all seatings of the pairs in a circle, and
let $E_i$ be the set of seatings in which pair i is together, for $1\le i\le m$.
Using Inclusion-Exclusion and the fact that n people can be seated in a circle in $(n-1)!$ ways,
$\displaystyle|\overline{E_1}\cap\cdots\cap\overline{E_m}|=|S|-\sum_{i}|E_i|+\sum_{i<j}|E_i\cap E_j|-\sum_{i<j<k}|E_i\cap E_j\cap E_k|+\cdots+(-1)^m|E_1\cap\cdots\cap E_m|$
$\;=(2m-1)!-\binom{m}{1}2^1(2m-2)!+\binom{m}{2}2^2(2m-3)!-\binom{m}{3}2^3(2m-4)!+\cdots+(-1)^m\binom{m}{m}2^m(m-1)!$
$\;\;=\displaystyle\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}2^{i}(2m-i-1)!$.
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct boxes?
You correctly found that $4$ indistinguishable oranges can be placed in $5$ distinct boxes in
$$\binom{4 + 5 - 1}{5 - 1} = \binom{8}{4}$$
ways. Since there are five ways each of the six apples can be placed in a box, there are $5^6$ ways to distribute the apples. Hence, the number of possible distributions is
$$5^6\binom{8}{4}$$
As for the other formula you considered, if you wanted to arrange four indistinguishable dividers and six different apples in a row, you could choose four of the ten positions for the dividers in $\binom{10}{4}$ ways, then arrange the six apples in the six remaining positions in $6!$ ways, which yields
$$\binom{10}{4}6! = \frac{10!}{4!6!} \cdot 6! = \frac{10!}{4!} = \frac{(6 + 5 - 1)!}{(5 - 1)!}$$
However, we do not care about the order of the apples within each box, so this formula yields too large a number for the number of ways of distributing the apples.
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct so that there are two pieces of fruit in each box?
The restriction that exactly two pieces of fruit are placed in each box means that we have three cases to consider:
- Two oranges apiece are placed in each of two boxes.
- Two oranges are placed in one box and one orange apiece is placed in each of two other boxes.
- Each orange is placed in a distinct box.
Two oranges apiece are placed in each of two boxes: Choose which two of the five boxes each receive two oranges. Choose which two of the six apples are placed in the leftmost empty box. Choose which two of the remaining four apples are placed in the leftmost remaining empty box. Place the final two apples in the remaining empty box.
There are $$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ such distributions.
Two oranges are placed in one box and one orange apiece is placed in each of two other boxes: Choose which of the five boxes receives two oranges. Choose which two of the remaining four boxes each receive one of the other two oranges. Choose which of the six apples is placed in the leftmost box with one apple and which of the remaining five apples is placed in the other box with one apple. Choose which two of the remaining four apples is placed in the leftmost empty box. The other two apples must be placed in the remaining empty box.
There are $$\binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$ such distributions.
Each orange is placed in a distinct box: There are $\binom{5}{4}$ ways to select which four of the boxes each receives one orange. There are $\binom{6}{2}$ ways to select which two apples are placed in the empty box. There are $4!$ ways to distribute the remaining four apples to the four boxes that each contain one orange.
There are $$\binom{5}{4}\binom{6}{2}4!$$ such distributions.
Since these cases are mutually exclusive and exhaustive, the number of ways to distribute four indistinguishable oranges and six distinct apples in such a way that two pieces of fruit are placed in each box is found by adding the results for each case.
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{6}{2}4!$$
Best Answer
I won't give you an actual answer, and this post is not a hint to solve the given question, but I will suggest a way that you can simplify the problem in order to understand better what's being asked.
(By the way, that's often a very useful approach, that is, simplifying the problem to get a better understanding of it)
Say you have a square and you assign numbers to the edges, in a clockwise manner, starting from the top edge. There are 4! = 24 different assignments you could make. Here are a few examples: 1234, 3241, 1324, 4123, 4321, and so on.
Now, note that the assignments 1234, 2341, 3412, and 4123 don't actually represent different squares because you could rotate them to match one another. You could put them into a grouping of their own. On the other hand, 1234 and 1324 are different assignments in the sense that no matter how you rotate either of the resulting squares, you won't be able to make them match. So, 1234 and 1324 must belong to different groupings.
What the question about the dice is asking is how many such groupings there are, but with a cube and its faces, instead of a square and its edges. Note that you can rotate a cube in more ways than you can rotate a square.
Edit: Based on the comment exchange below, here's the brute-force solution for the analogous problem with triangles. It's brute-force because I'm actually listing all the possible assignments and explicitly testing if they're pairwise equivalent or not. You will not want to do that in the dice problem. That's not how you want to solve that problem. I'm only brute-forcing the solution here to make it easier for you to understand what the problem is asking.
You can divide the 3! = 6 possible assignments into only 2 groups where every assignment in one group is equivalent to the other assignments in the same group but different from all assignments in the other group. The solution to "how many such groups are there?" is, then, 2. One is coloured blue, the other red in the figure below.
Hint #1 In the case of 6-sided dice, if all the dice have opposite sides adding up to 7 (1 opposite to 6, 2 opposite to 5, 3 opposite to 4) then there are only 2 groups. Try to prove that result, then look at the case when opposite faces don't have to add up to 7. How many of those cases are there in total?
Hint #2 So, if you fix the top and bottom numbers, you get 2 groups (that's the result of hint #1 above). In how many ways can you fix the top and bottom numbers?