In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?

Options

A. `(9!)/(4!3!2!) `
B. `(10!)/(4!3!2!) `
C. `(9!) `
D. None of these

Explanation

To find the number of ways to arrange 4 red, 3 yellow, and 2 green discs in a row, where the discs of the same color are indistinguishable, we can use the formula for permutations of multiset.

### Step-by-step Solution:

1. **Identify the Total Number of Discs**:
- We have 4 red discs, 3 yellow discs, and 2 green discs.
- Total number of discs = 4 (red) + 3 (yellow) + 2 (green) = 9 discs.

2. **Use the Permutation Formula for Indistinguishable Objects**:
- The formula for the number of arrangements of n objects where there are p1 indistinguishable objects of one type, p2 indistinguishable objects of another type, and so on, is given by:
\[ \text{Number of arrangements} = \frac{n!}{p1! \times p2! \times p3!} \]
- Here, \( n = 9 \), \( p1 = 4 \) (red), \( p2 = 3 \) (yellow), and \( p3 = 2 \) (green).

3. **Plug in the Values into the Formula**:
- Substitute the values into the formula:
\[ \text{Number of arrangements} = \frac{9!}{4! \times 3! \times 2!} \]

4. **Calculate Factorials**:
- Calculate \( 9! \):
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \]
- Calculate \( 4! \):
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
- Calculate \( 3! \):
\[ 3! = 3 \times 2 \times 1 = 6 \]
- Calculate \( 2! \):
\[ 2! = 2 \times 1 = 2 \]

5. **Substitute the Factorials Back into the Formula**:
- Now substitute the calculated factorials back into the formula:
\[ \text{Number of arrangements} = \frac{362880}{24 \times 6 \times 2} \]

6. **Calculate the Denominator**:
- Calculate \( 24 \times 6 = 144 \)
- Then calculate \( 144 \times 2 = 288 \)

7. **Final Calculation**:
- Now divide \( 362880 \) by \( 288 \):
\[ \text{Number of arrangements} = \frac{362880}{288} = 1260 \]

### Conclusion:
The total number of ways to arrange the discs is **1260**.