If the difference between the number of subsets of two sets A and B is 120, then `n(A xx B)` is equal to

Options

A. 21
B. 25
C. 18
D. 24

Answer

To solve the problem, we need to find the number of elements in the Cartesian product of two sets A and B, given that the difference between the number of subsets of these two sets is 120.

### Step-by-Step Solution:

1. **Understanding the Number of Subsets**:
The number of subsets of a set with \( n \) elements is given by \( 2^n \).
Therefore, if set A has \( p \) elements, the number of subsets of A is \( 2^p \), and if set B has \( q \) elements, the number of subsets of B is \( 2^q \).

2. **Setting Up the Equation**:
According to the problem, the difference between the number of subsets of sets A and B is 120:
\[ 2^p - 2^q = 120 \]

3. **Factoring Out Common Terms**:
We can factor out \( 2^q \) from the left side:
\[ 2^q (2^{p-q} - 1) = 120 \]

4. **Finding Possible Values**:
Now we need to find pairs of \( (2^q, 2^{p-q} - 1) \) that multiply to 120. We can start by factoring 120:
\[ 120 = 2^3 \times 3 \times 5 \]
The factors of 120 are: \( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 \).

5. **Testing Values for \( 2^q \)**:
Since \( 2^q \) must be a power of 2, the possible values for \( 2^q \) are \( 1, 2, 4, 8 \).

- If \( 2^q = 8 \) (which means \( q = 3 \)):
\[ 2^q (2^{p-q} - 1) = 120 \implies 8(2^{p-3} - 1) = 120 \implies 2^{p-3} - 1 = 15 \implies 2^{p-3} = 16 \implies p - 3 = 4 \implies p = 7 \]

6. **Finding the Number of Elements in \( A \times B \)**:
The number of elements in the Cartesian product \( A \times B \) is given by:
\[ n(A \times B) = n(A) \times n(B) = p \times q = 7 \times 3 = 21 \]

### Final Answer:
Thus, the number of elements in \( A \times B \) is \( 21 \).