To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we need to understand the concepts of permutations and combinations. ### Step 1: Understand the Definitions 1. **Permutations**: The number of ways to arrange \(r\) objects from \(n\) distinct objects is given by the formula: \[ ^{n}P_{r} = \frac{n!}{(n-r)!} \] For our case, where \(r = 3\): \[ ^{n}P_{3} = \frac{n!}{(n-3)!} \] 2. **Combinations**: The number of ways to choose \(r\) objects from \(n\) distinct objects is given by the formula: \[ ^{n}C_{r} = \frac{n!}{r!(n-r)!} \] For our case, where \(r = 4\): \[ ^{n}C_{4} = \frac{n!}{4!(n-4)!} \] ### Step 2: Substitute the Formulas into the Equation Now, substituting these formulas into the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\): \[ \frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0 \] ### Step 3: Simplify the Equation We can factor out \(n!\) from both terms: \[ n! \left(\frac{1}{(n-3)!} - 6 \cdot \frac{1}{4!(n-4)!}\right) = 0 \] Since \(n! \neq 0\) for \(n \geq 0\), we can simplify the equation to: \[ \frac{1}{(n-3)!} - 6 \cdot \frac{1}{4!(n-4)!} = 0 \] ### Step 4: Solve for \(n\) Now, we can rewrite \(4!\) as \(24\): \[ \frac{1}{(n-3)!} = \frac{6}{24(n-4)!} \] This simplifies to: \[ \frac{1}{(n-3)!} = \frac{1}{4(n-4)!} \] Cross-multiplying gives: \[ 4(n-4)! = (n-3)! \] Using the property of factorials, we know that: \[ (n-3)! = (n-3)(n-4)! \] Substituting this into the equation gives: \[ 4(n-4)! = (n-3)(n-4)! \] Assuming \(n-4! \neq 0\) (which is true for \(n \geq 4\)), we can divide both sides by \((n-4)!\): \[ 4 = n - 3 \] ### Step 5: Solve for \(n\) Now, solving for \(n\): \[ n = 4 + 3 = 7 \] ### Conclusion Thus, the value of \(n\) is \(7\). ### Verification of Options - **Option A (5)**: Incorrect, as substituting \(n = 5\) does not satisfy the original equation. - **Option B (6)**: Incorrect, as substituting \(n = 6\) does not satisfy the original equation. - **Option C (7)**: Correct, as we have shown that \(n = 7\) satisfies the equation. - **Option D (8)**: Incorrect, as substituting \(n = 8\) does not satisfy the original equation. ### Revision Summary - **Permutations** and **Combinations** have distinct formulas: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) and \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\). - Factor out common terms to simplify equations involving factorials. - Use properties of factorials to relate different factorial expressions. - Always verify your solution by substituting back into the original equation. The correct answer is **C. 7**.