Number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour.

Answer

To solve the problem of arranging 20 different pearls of two colors (10 pearls of each color) alternately on a necklace, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
- We have 10 pearls of red color (R) and 10 pearls of green color (G).
- The arrangement must be alternating, meaning we can have either RGRGRGRGRG... or GRGRGRGRGR...

2. **Fixing the First Pearl**:
- Since the arrangement is circular (like a necklace), we can fix one pearl to eliminate the effect of rotation. Let's fix one red pearl (R).
- This leaves us with 9 red pearls and 10 green pearls to arrange.

3. **Arranging the Remaining Pearls**:
- After fixing one red pearl, we have the following arrangement:
- R _ R _ R _ R _ R _ R _ R _ R _ R _ R _
- There are 9 gaps between the red pearls where we can place the green pearls.

4. **Placing the Green Pearls**:
- We need to place the 10 green pearls in the 10 available gaps (one gap after each red pearl and one gap before the first red pearl).
- The arrangement of the 10 green pearls can be done in \(10!\) ways.

5. **Arranging the Red Pearls**:
- The remaining 9 red pearls can be arranged in the remaining gaps. Since we have fixed one red pearl, we can arrange the remaining 9 red pearls in \(9!\) ways.

6. **Considering the Necklace Symmetry**:
- Since the necklace can be flipped, we need to divide the total arrangements by 2 to account for this symmetry.

7. **Calculating the Total Arrangements**:
- The total number of arrangements is given by:
\[ \text{Total Arrangements} = \frac{10! \times 9!}{2} \]

### Final Calculation:
- We can compute \(10!\) and \(9!\):
- \(10! = 3628800\)
- \(9! = 362880\)

Thus,
\[ \text{Total Arrangements} = \frac{3628800 \times 362880}{2} \]

### Conclusion:
The total number of ways to arrange the pearls alternately on the necklace is:
\[ \text{Total Arrangements} = \frac{1316818944000}{2} = 658409472000 \]