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.

What are your thoughts? What have you tried? What tools are available to you to solve the problem? Please edit the question to provide this information. Questions that show no evidence of any independent effort to reach a solution usually are not well received here. — Robert Shore, Jun 19 '23 at 05:18
I'll ask the second question again, with an example: Can you move a bead from one end to the other? How many distinct necklaces can two blue and one red bead make? Also, does "Burnside's lemma" sound familiar? — Arthur, Jun 19 '23 at 05:27
Does this answer your question? Necklace problem with Burnside Lemma — Dan Uznanski, Jun 20 '23 at 12:30
As to say how many eight-digit numbers can be formed with four digits 1, two digits 3 and two digits 5? — Ataulfo, Jun 20 '23 at 12:32
@Piquito That is a different question since the objects are arranged in a line. In this question, two arrangements are considered to be equivalent if they can be obtained from each other by rotation and/or reflection. — N. F. Taussig, Jun 21 '23 at 10:48

JMP · Accepted Answer · 2023-06-20T14:43:02.843

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.

this is missing several: there are 33 total. 7 are missing: bwwwrbwr wbwwrbwr wwbwrbwr wwwbwrbwr wwwbrbwr bwwrwbwr bwwrbwwr bwwrwwbr); you also have bwwbwrwr and its mirror wbwwbrwr both counted. — Dan Uznanski, Jun 20 '23 at 13:21

A circle permutation problem.

1 Answers1