If `n(AnnB)=20`, `n(AnnBnnC)=5`, then `n(Ann(B-C))` is

Options

A. `15`
B. `10`
C. `5`
D. `20`

Answer

To solve the problem step by step, we need to find \( n(A \cap (B - C)) \) given that \( n(A \cap B) = 20 \) and \( n(A \cap B \cap C) = 5 \).

### Step 1: Understand the given information
We know:
- \( n(A \cap B) = 20 \): This means there are 20 elements that are common to both sets A and B.
- \( n(A \cap B \cap C) = 5 \): This means there are 5 elements that are common to sets A, B, and C.

### Step 2: Identify \( n(A \cap (B - C)) \)
The expression \( A \cap (B - C) \) represents the elements that are in both A and B but not in C. We can express this as:
\[ n(A \cap (B - C)) = n(A \cap B) - n(A \cap B \cap C) \]

### Step 3: Substitute the known values
Now we can substitute the known values into the equation:
\[ n(A \cap (B - C)) = n(A \cap B) - n(A \cap B \cap C) \]
\[ n(A \cap (B - C)) = 20 - 5 \]

### Step 4: Calculate the result
Now, perform the subtraction:
\[ n(A \cap (B - C)) = 15 \]

### Final Answer
Thus, the value of \( n(A \cap (B - C)) \) is \( 15 \).

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