If `P(n,6) = 30. P(n,4)` then `n = ?`

Options

A. 10
B. 9
C. 8
D. None of these

Answer

To solve the equation \( P(n, 6) = 30 \cdot P(n, 4) \), we will use the formula for permutations, which states:

\[ P(n, k) = \frac{n!}{(n-k)!} \]

### Step-by-Step Solution:

1. **Write the permutation expressions**:
\[ P(n, 6) = \frac{n!}{(n-6)!} \]
\[ P(n, 4) = \frac{n!}{(n-4)!} \]

2. **Substitute into the equation**:
\[ \frac{n!}{(n-6)!} = 30 \cdot \frac{n!}{(n-4)!} \]

3. **Cancel \( n! \) from both sides** (assuming \( n! \neq 0 \)):
\[ \frac{1}{(n-6)!} = 30 \cdot \frac{1}{(n-4)!} \]

4. **Rewrite the right-hand side**:
\[ \frac{1}{(n-6)!} = \frac{30}{(n-4)(n-5)(n-6)!} \]

5. **Multiply both sides by \( (n-6)! \)**:
\[ 1 = \frac{30}{(n-4)(n-5)} \]

6. **Cross-multiply**:
\[ (n-4)(n-5) = 30 \]

7. **Expand the left-hand side**:
\[ n^2 - 9n + 20 = 30 \]

8. **Rearrange the equation**:
\[ n^2 - 9n - 10 = 0 \]

9. **Factor the quadratic equation**:
\[ (n - 10)(n + 1) = 0 \]

10. **Solve for \( n \)**:
\[ n - 10 = 0 \quad \Rightarrow \quad n = 10 \]
\[ n + 1 = 0 \quad \Rightarrow \quad n = -1 \]

11. **Determine valid solutions**:
Since \( n \) must be a non-negative integer (as factorials for negative integers do not exist), we discard \( n = -1 \).

Thus, the final answer is:
\[ \boxed{10} \]