If `P_(n)` denotes the product of all the coefficients of `(1+x)^(n)` and `9!P_(n+1)=10^(9)P_(n)` then `n` is equal to

Options

A. `10`
B. `9`
C. `19`
D. none of these

Explanation

To solve the problem, we need to find the value of \( n \) such that \( 9! P_{n+1} = 10^9 P_n \), where \( P_n \) denotes the product of all coefficients of the expansion of \( (1+x)^n \).

### Step-by-Step Solution:

1. **Understanding \( P_n \)**:
The coefficients of the expansion of \( (1+x)^n \) are given by \( \binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n} \). Therefore, the product of all coefficients \( P_n \) can be expressed as:
\[ P_n = \binom{n}{0} \cdot \binom{n}{1} \cdot \binom{n}{2} \cdots \binom{n}{n} \]

2. **Expressing \( P_{n+1} \)**:
Similarly, for \( n+1 \):
\[ P_{n+1} = \binom{n+1}{0} \cdot \binom{n+1}{1} \cdot \binom{n+1}{2} \cdots \binom{n+1}{n+1} \]

3. **Finding the ratio \( \frac{P_{n+1}}{P_n} \)**:
We need to compute \( \frac{P_{n+1}}{P_n} \):
\[ \frac{P_{n+1}}{P_n} = \frac{\binom{n+1}{0} \cdot \binom{n+1}{1} \cdots \binom{n+1}{n}}{\binom{n}{0} \cdot \binom{n}{1} \cdots \binom{n}{n}} = \frac{1 \cdot \binom{n+1}{1} \cdots \binom{n+1}{n}}{1 \cdot \binom{n}{1} \cdots \binom{n}{n}} \]

4. **Calculating the individual terms**:
Each term can be calculated as follows:
\[ \frac{\binom{n+1}{k}}{\binom{n}{k}} = \frac{(n+1)!/(k!(n+1-k)!)}{n!/(k!(n-k)!)} = \frac{(n+1)!}{n!} \cdot \frac{(n-k)!}{(n+1-k)!} = \frac{n+1}{n+1-k} \]

5. **Combining the terms**:
Therefore, we can express the product:
\[ \frac{P_{n+1}}{P_n} = \prod_{k=0}^{n} \frac{n+1}{n+1-k} = \frac{(n+1)^{n+1}}{n!} \]

6. **Setting up the equation**:
Now substituting this back into the equation \( 9! P_{n+1} = 10^9 P_n \):
\[ 9! \cdot \frac{(n+1)^{n+1}}{n!} = 10^9 \]

7. **Simplifying**:
This can be rearranged to:
\[ \frac{(n+1)^{n+1}}{n!} = \frac{10^9}{9!} \]

8. **Finding \( n \)**:
Now we can try different integer values for \( n \) to see which satisfies the equation. Testing \( n = 9 \):
\[ P_9 = \binom{9}{0} \cdot \binom{9}{1} \cdots \binom{9}{9} = 1 \cdot 9 \cdot 36 \cdots \cdot 1 = 9! \text{ (as it includes all combinations)} \]
\[ P_{10} = \binom{10}{0} \cdot \binom{10}{1} \cdots \binom{10}{10} = 10! \]
Thus,
\[ 9! \cdot 10! = 10^9 \cdot 9! \]

This simplifies to \( 10! = 10^9 \), which is true when \( n = 9 \).

### Conclusion:
The value of \( n \) is \( \boxed{9} \).