In Example 28 , the number of ways of choosing A and B such that A = B, is

Options

A. `2^(n)`
B. `3^(n)`
C. `""^(2n)C_(n)`
D. none of these

Explanation

To solve the problem of finding the number of ways of choosing sets A and B such that A = B, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Sets**: Let set A contain R elements, where R can range from 0 to N (the total number of elements available). Therefore, we have \( 0 \leq R \leq N \).

2. **Choosing Set A**: The number of ways to choose R elements from N elements is given by the binomial coefficient \( \binom{N}{R} \). This represents the number of combinations of N elements taken R at a time.

\[ \text{Ways to choose A} = \binom{N}{R} \]

3. **Choosing Set B**: Since we want A to be equal to B, the subset B can have at most R elements. The number of ways to choose a subset B from A (which has R elements) is given by \( 2^R \). This is because each element in A can either be included in B or not, leading to \( 2^R \) possible combinations.

\[ \text{Ways to choose B} = 2^R \]

4. **Total Ways of Choosing A and B**: The total number of ways to choose both sets A and B such that A = B is the product of the ways to choose A and the ways to choose B.

\[ \text{Total ways} = \binom{N}{R} \cdot 2^R \]

5. **Summing Over All Possible R**: Since R can vary from 0 to N, we need to sum the total ways over all possible values of R.

\[ \text{Total ways} = \sum_{R=0}^{N} \binom{N}{R} \cdot 2^R \]

6. **Using the Binomial Theorem**: According to the binomial theorem, the sum \( \sum_{R=0}^{N} \binom{N}{R} x^R = (1 + x)^N \). If we set \( x = 2 \), we get:

\[ \sum_{R=0}^{N} \binom{N}{R} \cdot 2^R = (1 + 2)^N = 3^N \]

7. **Final Result**: Therefore, the total number of ways of choosing A and B such that A = B is:

\[ \text{Total ways} = 3^N \]

### Conclusion:
The number of ways of choosing A and B such that A = B is \( 3^N \).