In Example 28, the number of ways of choosing A
and B such that B contains just one element more then A, is

Options

A. `""^(2n)C_(n-1)`
B. `3^(n)`
C. `(2^(n))^(2)`
D. `""^(2n)C_(n)`

Explanation

To solve the problem of finding the number of ways to choose sets A and B such that B contains just one element more than A, we can follow these steps:

### Step 1: Define the Elements in Sets A and B
Let the number of elements in set A be denoted as \( R \). Since set B contains just one element more than A, the number of elements in set B will be \( R + 1 \).

### Step 2: Determine the Total Number of Elements
Assume we have a total of \( N \) elements available to choose from.

### Step 3: Calculate the Number of Ways to Choose Set A
The number of ways to choose \( R \) elements from \( N \) elements is given by the binomial coefficient:
\[ \binom{N}{R} \]

### Step 4: Calculate the Number of Ways to Choose Set B
Since set B has \( R + 1 \) elements, the number of ways to choose \( R + 1 \) elements from the \( N \) elements is:
\[ \binom{N}{R + 1} \]

### Step 5: Combine the Choices for A and B
The total number of ways to choose sets A and B is the product of the two choices:
\[ \text{Total Ways} = \binom{N}{R} \times \binom{N}{R + 1} \]

### Step 6: Sum Over All Possible Values of R
Since \( R \) can vary from \( 0 \) to \( N - 1 \), we need to sum the total ways for all values of \( R \):
\[ \text{Total Ways} = \sum_{R=0}^{N-1} \binom{N}{R} \times \binom{N}{R + 1} \]

### Step 7: Simplify the Summation
Using the identity that relates the sum of products of binomial coefficients, we find that:
\[ \sum_{R=0}^{N-1} \binom{N}{R} \times \binom{N}{R + 1} = 2^N \cdot \binom{N}{N - 1} \]
This simplifies to:
\[ = 2^N \]

### Final Result
Thus, the total number of ways of choosing sets A and B such that B contains just one element more than A is:
\[ 2^N \]