Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye ?

Options

A. 3255
B. 212
C. 3720
D. none of these

Answer

To solve the problem of how many combinations of dyes can be chosen taking at least one green and one blue dye from the given dyes, we can break down the solution into clear steps:

### Step 1: Calculate the combinations of green dyes
We have 5 different green dyes. The total number of ways to choose any combination of these dyes (including choosing none) is given by \(2^n\), where \(n\) is the number of items. Thus, for 5 green dyes:
\[ \text{Total combinations of green dyes} = 2^5 = 32 \]
However, we need at least one green dye, so we subtract the case where no green dye is chosen:
\[ \text{At least one green dye} = 32 - 1 = 31 \]

### Step 2: Calculate the combinations of blue dyes
Similarly, we have 4 different blue dyes. The total number of ways to choose any combination of these dyes is:
\[ \text{Total combinations of blue dyes} = 2^4 = 16 \]
Again, we need at least one blue dye, so we subtract the case where no blue dye is chosen:
\[ \text{At least one blue dye} = 16 - 1 = 15 \]

### Step 3: Calculate the combinations of red dyes
For the 3 different red dyes, there are no restrictions on how many can be chosen (including choosing none). Thus, the total number of combinations is:
\[ \text{Total combinations of red dyes} = 2^3 = 8 \]

### Step 4: Combine the results
Now, to find the total number of combinations that include at least one green dye and at least one blue dye, we multiply the number of ways to choose green dyes, blue dyes, and red dyes:
\[ \text{Total combinations} = (\text{At least one green dye}) \times (\text{At least one blue dye}) \times (\text{Total combinations of red dyes}) \]
Substituting the values we calculated:
\[ \text{Total combinations} = 31 \times 15 \times 8 \]

### Step 5: Calculate the final result
Now, we perform the multiplication:
\[ 31 \times 15 = 465 \]
Then, multiply by 8:
\[ 465 \times 8 = 3720 \]

Thus, the total number of combinations of dyes that can be chosen taking at least one green and one blue dye is **3720**.

### Final Answer:
The answer is **3720**.

---