A counting problem using Burnside's lemma Suppose we have 12 objects (say, 6 indistinguishable black ones and 6 indistinguishable white ones). How many seatings at a round table can we form from them?
The answer is $80$, but how could this be counted via Burnside's lemma?
 A: It might be a good idea to try the problem for a smaller number of objects as a warmup.  For example, with two black and two white objects, there are two distinct seatings up to rotation of the table: BBWW and BWBW; with three black and three white, there are four: BBBWWW, BBWBWW, BBWWBW, and BWBWBW.  (Do these answers agree with your method?)  Burnside's Lemma says that these numbers are equal to the average over the relevant group ($C_4$ or $C_6$) of the number of seatings that are fixed by a group element.
To compute the number of fixed seatings, it helps to write out the group elements using cycle notation.  The elements of $C_4$ are
$$(1)(2)(3)(4),\quad(1234),\quad(13)(24),\quad(1432).$$
For a seating to be fixed under one of these permutations, the cycles have to be single-colored.  For example, under $(13)(24),$ which is rotation by two seats, the seating won't be fixed unless seats $1$ and $3$ have the same color and seats $2$ and $4$ have the same color.  There are therefore two fixed seatings, BWBW and WBWB.  Under $(1234),$ which is rotation by one seat, all four seats would have to have the same color, which is impossible, so there are no fixed seatings.  It is similar for $(1432).$  Under the identity permutation, $(1)(2)(3)(4),$ all $\binom{4}{2}=6$ seatings are fixed.  Hence there are
$$\frac{1}{4}\left(6+0+2+0\right)=2$$
distinct seatings up to rotation.
The elements of $C_6$ are
$$(1)(2)(3)(4)(5)(6),\quad(123456),\quad(135)(246),\quad(14)(25)(36),\quad(153)(264),\quad(16543).$$
You can check that application of Burnside's Lemma yields
$$\frac{1}{6}\left(20+0+2+0+2+0\right)=4$$
distinct seatings up to rotation.
Now you can try the extend these ideas to larger numbers of objects.
A: Hint: Given a random arrangement of $W_i$ and $B_i$, when would the arrangement appear again? This can only happen if we rotate the table.
Hint: There is no arrangement which happens again after rotating once.
There are arrangements which happen again after rotating 2 or 4 times. E.g. $W_1B_1W_2B_2\ldots W_6 B_6$.
Are there arrangements which happen after rotating 3 times?
Now apply Burnside's Lemma.
