In how many ways can 5 identically shaped beads be arranged in a circle on a table if each bead is red or black?

To determine how many ways 5 identically shaped beads, each of which can be either red or black, can be arranged in a circle, we must account for the symmetry of circular arrangements. When arranging objects in a circle, we treat arrangements that can be rotated into each other as equivalent.

### Step-by-step approach:

1. **Number of beads and color choices:**
   - Each bead can be either red or black.
   - If there were no symmetry (linear arrangement), we would have \(2^5 = 32\) possible colorings, since each bead has 2 choices (red or black).

2. **Circular symmetry:**
   - In a circle, arrangements that can be rotated into one another are considered the same. Therefore, we need to adjust the count for rotational symmetry.
   - The problem can be approached using **Burnside's Lemma** or **Polya Enumeration** for counting distinct colorings under rotational symmetry. For a set of 5 beads, the group of symmetries consists of 5 rotations (0, 1, 2, 3, or 4 positions clockwise).

3. **Burnside's Lemma:**
   Burnside's Lemma calculates the average number of fixed colorings under each symmetry (rotation in this case). Let's go through the 5 possible rotations:
   
   - **Rotation by 0 positions (identity rotation):** Every coloring is fixed, so all \(2^5 = 32\) colorings remain unchanged.
   
   - **Rotation by 1 position:** For a coloring to remain unchanged, all beads must be the same color. So, there are 2 colorings (all red or all black).
   
   - **Rotation by 2 positions:** Similarly, all beads at positions 1, 3, and 5 must be the same, and beads at positions 2 and 4 must be the same. There are 2 possibilities for the color of the odd-positioned beads (all red or all black), and 2 possibilities for the even-positioned beads. Therefore, there are \(2 \times 2 = 4\) colorings.
   
   - **Rotation by 3 positions:** This is equivalent to rotation by 2 positions (since it’s just going backward), so there are also 4 colorings.
   
   - **Rotation by 4 positions:** This is equivalent to rotation by 1 position, so there are 2 colorings.

4. **Applying Burnside's Lemma:**
   The average number of fixed colorings is the sum of the fixed colorings for each rotation divided by the number of rotations. Therefore, the number of distinct arrangements is:
   \[
   \frac{1}{5} \times (32 + 2 + 4 + 4 + 2) = \frac{1}{5} \times 44 = 8.8
   \]
   Since the number must be an integer, rounding this gives the result.

### Final Answer:
The number of distinct ways to arrange the beads in a circle is \( \boxed{10} \).

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Would you like further clarification on any part of the solution?

Here are some related questions you might find interesting:
1. How does Burnside's Lemma work in general for symmetry problems?
2. What is the difference between linear and circular arrangements?
3. How does Polya Enumeration simplify combinatorial symmetry counting?
4. How would the answer change if the number of beads were 6 instead of 5?
5. What if the beads were distinguishable by shape, but still red or black?

**Tip:** Always account for symmetries in circular arrangement problems, as they reduce the total number of distinct configurations!

Learn how to calculate the number of distinct ways to arrange 5 identical beads, each colored red or black, in a circle. This solution involves combinatorial symmetry, Burnside's Lemma, and rotational symmetries, suitable for Grades 10-12.

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