If a denotes the number of permutation of x+2 things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x -11 things taken all at a time such that a=182bc, then the value of x will be

Options

A. A. 12
B. B. -12
C. C. 6
D. D. none of these

Answer

To solve the problem, we need to find the value of \( x \) given the relationships between permutations \( a \), \( b \), and \( c \).

1. **Define the permutations**:
- \( a \) is the number of permutations of \( x + 2 \) things taken all at a time, which can be expressed as:
\[ a = (x + 2)! \]
- \( b \) is the number of permutations of \( x \) things taken 11 at a time:
\[ b = P(x, 11) = \frac{x!}{(x - 11)!} \]
- \( c \) is the number of permutations of \( x - 11 \) things taken all at a time:
\[ c = (x - 11)! \]

2. **Set up the equation**:
According to the problem, we have:
\[ a = 182bc \]
Substituting the expressions for \( a \), \( b \), and \( c \):
\[ (x + 2)! = 182 \left( \frac{x!}{(x - 11)!} \right) (x - 11)! \]
Simplifying this gives:
\[ (x + 2)! = 182 x! \]

3. **Express \( (x + 2)! \)**:
We can express \( (x + 2)! \) in terms of \( x! \):
\[ (x + 2)! = (x + 2)(x + 1)(x!) \]
Thus, we can rewrite the equation:
\[ (x + 2)(x + 1)(x!) = 182 x! \]

4. **Cancel \( x! \)** (assuming \( x! \neq 0 \)):
\[ (x + 2)(x + 1) = 182 \]

5. **Expand and rearrange the equation**:
\[ x^2 + 3x + 2 = 182 \]
\[ x^2 + 3x - 180 = 0 \]

6. **Factor the quadratic equation**:
We look for two numbers that multiply to \(-180\) and add to \(3\). The numbers \(15\) and \(-12\) work:
\[ (x + 15)(x - 12) = 0 \]

7. **Solve for \( x \)**:
Setting each factor to zero gives:
\[ x + 15 = 0 \quad \Rightarrow \quad x = -15 \quad (\text{not valid since } x \text{ must be non-negative}) \]
\[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \]

Thus, the value of \( x \) is \( \boxed{12} \).