To solve the equation \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), we start by using the formulas for permutations and combinations: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) \(^{n}P_{3} = \frac{n!}{(n-3)!}\) \(^{n}C_{4} = \frac{n!}{4!(n-4)!}\) \(\frac{n!}{(n-3)!} - 6 \left(\frac{n!}{4!(n-4)!}\right) = 0\) Factor out \(n!\) Factoring \(n!\) from both terms gives: \(n! \left(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!}\right) = 0\) Since \(n! \neq 0\), we can simplify to: \(\frac{1}{(n-3)!} - \frac{6}{4!(n-4)!} = 0\) Recall that \(4! = 24\), so: \(\frac{6}{4!(n-4)!} = \frac{6}{24(n-4)!} = \frac{1}{4(n-4)!}\) The equation now is: \(\frac{1}{(n-3)!} - \frac{1}{4(n-4)!} = 0\) Multiply through by \(4(n-3)!(n-4)!\) gives: \(4(n-4)! - (n-3)! = 0\) Recall that \((n-3)! = (n-3)(n-4)!\), so: \(4(n-4)! - (n-3)(n-4)! = 0\) \((n-4)!\) Factoring out \((n-4)!\) gives: \((n-4)! \left(4 - (n-3)\right) = 0\) Since \((n-4)! \neq 0\), we have: \(4 - (n-3) = 0\) Solving for \(n\): \(n - 3 = 4 \implies n = 7\) The value of n is: 7