Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combination of dyes which can be chosen taking at least one green and one blue dye is

Options

A. 3600
B. 3720
C. 3800
D. none of these

Answer

To solve the problem of finding the number of combinations of dyes that can be chosen while taking at least one green dye and one blue dye, we can break it down into manageable steps.

### Step-by-Step Solution:

1. **Identify the Dyes**:
- We have 5 different green dyes.
- We have 4 different blue dyes.
- We have 3 different red dyes.

2. **Calculate Combinations for Green Dyes**:
- We need to select at least one green dye from the 5 available. The number of ways to select at least one green dye can be calculated using the formula for combinations:
\[ \text{Total combinations for green} = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 \]
- Calculating each term:
- \(5C1 = 5\)
- \(5C2 = 10\)
- \(5C3 = 10\)
- \(5C4 = 5\)
- \(5C5 = 1\)
- Adding these together:
\[ 5 + 10 + 10 + 5 + 1 = 31 \]

3. **Calculate Combinations for Blue Dyes**:
- Similarly, we need to select at least one blue dye from the 4 available:
\[ \text{Total combinations for blue} = 4C1 + 4C2 + 4C3 + 4C4 \]
- Calculating each term:
- \(4C1 = 4\)
- \(4C2 = 6\)
- \(4C3 = 4\)
- \(4C4 = 1\)
- Adding these together:
\[ 4 + 6 + 4 + 1 = 15 \]

4. **Calculate Combinations for Red Dyes**:
- There are no restrictions on red dyes, so we can choose any number from 0 to 3:
\[ \text{Total combinations for red} = 3C0 + 3C1 + 3C2 + 3C3 \]
- Calculating each term:
- \(3C0 = 1\)
- \(3C1 = 3\)
- \(3C2 = 3\)
- \(3C3 = 1\)
- Adding these together:
\[ 1 + 3 + 3 + 1 = 8 \]

5. **Calculate Total Combinations**:
- Now, we multiply the combinations of green, blue, and red dyes:
\[ \text{Total combinations} = (\text{Combinations for green}) \times (\text{Combinations for blue}) \times (\text{Combinations for red}) \]
- Substituting the values:
\[ \text{Total combinations} = 31 \times 15 \times 8 \]
- First, calculate \(31 \times 15\):
\[ 31 \times 15 = 465 \]
- Then, multiply by 8:
\[ 465 \times 8 = 3720 \]

### Final Answer:
The total number of combinations of dyes which can be chosen taking at least one green dye and one blue dye is **3720**.