Discrete math:Placing beads on a necklace

[hsetennis]

Homework Statement

There are 8 beads: 4 black, 3 white, and 1 red.
How many ways can these be arranged on necklace?

Homework Equations

Just combinations nCm = (n-m)!/(m!).

The Attempt at a Solution

8C4 = 70, 4C3 = 24, and 1C1 = 1, so the total number of combinations is 95?

Or do I multiply them, resulting in 1680 possibilities?
Multiplication seems like the logical way of solving it, because it is a union of the sets of possibilities, but I haven't done this is a long time so I'm unsure.

 

[HallsofIvy]

The point of the "necklace" is that you are placing them circularly, not linearly- there are no "end" beads. So choose anyone bead you want, the red one, say, and arrange the others on that. If you have 7 beads, 4 black and 3 white, in how many ways can you arrange them?

 

[hsetennis]

The point of the "necklace" is that you are placing them circularly, not linearly- there are no "end" beads. So choose anyone bead you want, the red one, say, and arrange the others on that. If you have 7 beads, 4 black and 3 white, in how many ways can you arrange them?

Okay, so if I fix the red bead, then there would be 7C3 = 7C4 = 7*6*5/3/2 = 35. Then 35*8 for each choice of the red bead is 280 total. Is this correct?

For context, this is a smaller problem in a larger problem that I'm working on for Polya-Burnside Enumeration.

 

[haruspex]

Okay, so if I fix the red bead, then there would be 7C3 = 7C4 = 7*6*5/3/2 = 35. Then 35*8 for each choice of the red bead is 280 total. Is this correct?

You're right to take the red bead as a point of reference, but there's no reason to multiply by the number of positions it can occupy at the end. Those arrangements are all the same, as far as the necklace as an entity is concerned. Indeed, you need to remove some arrangements that you have counted twice (reflections).

 

[hsetennis]

Thanks Haruspex, I'm accounting for the reflections and rotations D8 with the Polya-Burnside lemma