If A and B are two sets such that A `cup` B has 20 elements , A has 8 elements and B has 16 elements , how many elements does `Acap` B have ?

Answer

To solve the problem, we need to find the number of elements in the intersection of sets A and B, denoted as \( A \cap B \). We can use the formula for the union of two sets:

\[ N(A \cup B) = N(A) + N(B) - N(A \cap B) \]

Where:
- \( N(A \cup B) \) is the number of elements in the union of sets A and B,
- \( N(A) \) is the number of elements in set A,
- \( N(B) \) is the number of elements in set B,
- \( N(A \cap B) \) is the number of elements in the intersection of sets A and B.

### Step-by-step solution:

1. **Identify the given values:**
- \( N(A \cup B) = 20 \) (the number of elements in the union of A and B)
- \( N(A) = 8 \) (the number of elements in set A)
- \( N(B) = 16 \) (the number of elements in set B)

2. **Substitute the values into the formula:**
\[ 20 = 8 + 16 - N(A \cap B) \]

3. **Simplify the equation:**
- First, calculate \( 8 + 16 \):
\[ 8 + 16 = 24 \]
- Now substitute back into the equation:
\[ 20 = 24 - N(A \cap B) \]

4. **Rearrange the equation to solve for \( N(A \cap B) \):**
- Move \( N(A \cap B) \) to one side and 20 to the other:
\[ N(A \cap B) = 24 - 20 \]

5. **Calculate \( N(A \cap B) \):**
\[ N(A \cap B) = 4 \]

### Final Answer:
The number of elements in \( A \cap B \) is 4.

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