A circle permutation problem.

Question

Four white beads, two red beads, and two blue beads are strung together to form a necklace. How many ways are there to string the beads?

By the way, the beads are indistinguishable.

Respond to @Arthur: Assume flip a necklace over and still have the "same" necklace? Eg. two red beads, and two blue beads can be strung into two kinds of necklace.

Thanks for @Robert Shore. My thought is to order all heads first, then cancel the order of beads of the same color. like $\frac{A_8^8}{A_4^4A_2^2A_2^2*}$. But clearly, there is a symmetric issue that should be considered like @Arthur pointed out.

Answer 1

Start with $4$ white and $2$ red beads. There are $3$ bracelets (necklaces where flips are equivalent).

W W W W R R
W W W R W R
W W R W W R

Now position the $2$ blue beads, looking at each case separately.

There are at most $28$ positions for the $2$ beads over each necklace, but this estimate is far too large.

B B W W W W R R
B W B W W W R R
B W W B W W R R
B W W W B W R R
B W W W W B R R
W B B W W W R R
W B W B W W R R
W B W W B W R R
W W B B W W R R

B W W W W R B R
W B W W W R B R
W W B W W R B R

W W W W R B B R

13, and

B B W W W R W R
B W B W W R W R
B W W B W R W R
B W W W B R W R
W B B W W R W R
W B W B W R W R

B W W W R B W R
W B W W R B W R
W W B W R B W R
W W W B R B W R

W W W R B B W R
W W W R B W B R

12, and

B B W W R W W R
B W B W R W W R
B W W B R W W R
W B B W R W W R

W B W R W B W R
W W B R W B W R
W W B R B W W R
W W B R W W B R

8, for a total of 33.