If `p=""^(n+2) P_(n+2), q=""^n P_11, r="^(n-11) P_(n-11)` and if p=182 qr then the value of n is

Options

A. 10
B. 12
C. 15
D. 18

Answer

To solve the problem, we need to find the value of \( n \) given the equations for \( p \), \( q \), and \( r \) and the relationship \( p = 182 \cdot q \cdot r \).

### Step-by-Step Solution:

1. **Define the expressions for \( p \), \( q \), and \( r \)**:
\[ p = (n + 2) P (n + 2) = \frac{(n + 2)!}{(n + 2 - (n + 2))!} = (n + 2)! \]
\[ q = n P 11 = \frac{n!}{(n - 11)!} \]
\[ r = (n - 11) P (n - 11) = \frac{(n - 11)!}{(n - 11 - (n - 11))!} = (n - 11)! \]

2. **Substituting into the equation**:
Given \( p = 182 \cdot q \cdot r \), we can substitute the expressions:
\[ (n + 2)! = 182 \cdot \left(\frac{n!}{(n - 11)!}\right) \cdot (n - 11)! \]

3. **Simplifying the equation**:
The \( (n - 11)! \) terms cancel out:
\[ (n + 2)! = 182 \cdot n! \]

4. **Expanding \( (n + 2)! \)**:
We can express \( (n + 2)! \) as:
\[ (n + 2)! = (n + 2)(n + 1)n! \]
Thus, we have:
\[ (n + 2)(n + 1)n! = 182 \cdot n! \]

5. **Canceling \( n! \) from both sides** (assuming \( n! \neq 0 \)):
\[ (n + 2)(n + 1) = 182 \]

6. **Expanding the left-hand side**:
\[ n^2 + 3n + 2 = 182 \]

7. **Rearranging the equation**:
\[ n^2 + 3n + 2 - 182 = 0 \implies n^2 + 3n - 180 = 0 \]

8. **Factoring the quadratic equation**:
We need to find two numbers that multiply to \(-180\) and add to \(3\). The factors are \(15\) and \(-12\):
\[ (n + 15)(n - 12) = 0 \]

9. **Finding the values of \( n \)**:
Setting each factor to zero gives:
\[ n + 15 = 0 \implies n = -15 \quad \text{(not valid)} \]
\[ n - 12 = 0 \implies n = 12 \quad \text{(valid)} \]

10. **Conclusion**:
The value of \( n \) is \( 12 \).