If `n (xi)=50, n (A)=38, n(B)=30` then the least value of `n(AnnB)` is (i) 30 (ii) 38 (iii) 50 (iv) 18

Options

A. 30
B. 38
C. 50
D. 18

Answer

To find the least value of \( n(A \cap B) \) given \( n(\xi) = 50 \), \( n(A) = 38 \), and \( n(B) = 30 \), we can follow these steps:

### Step 1: Understand the relationship between sets
We know that the number of elements in the union of two sets \( A \) and \( B \) can be expressed as:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
This equation helps us relate the sizes of the sets and their intersection.

### Step 2: Set up the inequality
Since \( A \cup B \) is a subset of the universal set \( \xi \), we have:
\[ n(A \cup B) \leq n(\xi) = 50 \]
Substituting the expression for \( n(A \cup B) \) into this inequality gives:
\[ n(A) + n(B) - n(A \cap B) \leq 50 \]

### Step 3: Substitute the known values
Now, substitute the values of \( n(A) \) and \( n(B) \):
\[ 38 + 30 - n(A \cap B) \leq 50 \]
This simplifies to:
\[ 68 - n(A \cap B) \leq 50 \]

### Step 4: Rearrange the inequality
Rearranging the inequality, we get:
\[ -n(A \cap B) \leq 50 - 68 \]
\[ -n(A \cap B) \leq -18 \]
Multiplying through by -1 (and reversing the inequality) gives:
\[ n(A \cap B) \geq 18 \]

### Step 5: Conclusion
Thus, the least value of \( n(A \cap B) \) is:
\[ n(A \cap B) \geq 18 \]
Therefore, the least value of \( n(A \cap B) \) is 18, which corresponds to option (iv).

### Final Answer:
The least value of \( n(A \cap B) \) is **18**.

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