For a positive integer n if the mean of the
binomial coefficients in the expansion of `(a + b)^(2n - 3)` is 16 Then n is equal to

Options

A. 4
B. 5
C. 7
D. 9

Explanation

To solve the problem, we need to find the value of \( n \) given that the mean of the binomial coefficients in the expansion of \( (a + b)^{2n - 3} \) is 16.

### Step-by-step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients in the expansion of \( (a + b)^{2n - 3} \) are given by:
\[ \binom{2n - 3}{0}, \binom{2n - 3}{1}, \ldots, \binom{2n - 3}{2n - 3} \]
There are \( 2n - 3 + 1 = 2n - 2 \) coefficients in total.

2. **Calculating the Mean**:
The mean of these coefficients can be calculated as:
\[ \text{Mean} = \frac{\sum_{k=0}^{2n-3} \binom{2n-3}{k}}{2n - 2} \]
From the Binomial Theorem, we know that:
\[ \sum_{k=0}^{m} \binom{m}{k} = 2^m \]
Therefore, for \( m = 2n - 3 \):
\[ \sum_{k=0}^{2n-3} \binom{2n-3}{k} = 2^{2n - 3} \]
Thus, the mean becomes:
\[ \text{Mean} = \frac{2^{2n - 3}}{2n - 2} \]

3. **Setting Up the Equation**:
We know from the problem that this mean is equal to 16:
\[ \frac{2^{2n - 3}}{2n - 2} = 16 \]

4. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 2^{2n - 3} = 16(2n - 2) \]

5. **Substituting 16**:
Since \( 16 = 2^4 \), we can rewrite the equation:
\[ 2^{2n - 3} = 2^4(2n - 2) \]
This simplifies to:
\[ 2^{2n - 3} = 2^{4} \cdot (2n - 2) \]

6. **Equating the Powers of 2**:
We can express \( 2^{2n - 3} \) as:
\[ 2^{2n - 3} = 2^{4 + 1} \cdot (2n - 2) \implies 2n - 2 = 2^{2n - 7} \]

7. **Rearranging**:
Rearranging gives:
\[ 2n - 2 = 2^{2n - 7} \]

8. **Finding n**:
We can solve for \( n \) by substituting values. Testing \( n = 5 \):
\[ 2(5) - 2 = 10 - 2 = 8 \]
And checking:
\[ 2^{2(5) - 7} = 2^{10 - 7} = 2^3 = 8 \]
This holds true.

Thus, the value of \( n \) is:
\[ \boxed{5} \]