If \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\), find the value of x.

The formula for permutations is: \(^{n}P_{r} = \frac{n!}{(n-r)!}\) The formula for combinations is: \(^{n}C_{r} = \frac{n!}{r!(n-r)!}\) We are given that: \(\frac{^{8}P_{x}}{^{8}C_{x}} = 6\) Substituting the formulas for permutations and combinations, we get: \(\frac{\frac{8!}{(8-x)!}}{\frac{8!}{x!(8-x)!}} = 6\) Simplifying, we get: \(\frac{x!}{(8-x)!} = \frac{1}{6}\) Multiplying both sides by \(\frac{(8-x)!}{x!}\), we get: \(\frac{8!}{x!(8-x)!} = 6\) Dividing both sides by 8!, we get: \(\frac{1}{^{8}C_{x}} = \frac{1}{720}\) Multiplying both sides by 720, we get: \(^{8}C_{x} = 720\) We can use the formula for combinations to find the value of x: \(^{8}C_{x} = \frac{8!}{x!(8-x)!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!}\) Since \(^{8}C_{x} = 720\), we have: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{x!(8-x)!} = 720\) Dividing both sides by 720, we get: \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{720} = x!(8-x)!\) Simplifying, we get: \(x!(8-x)! = 40320\) The only value of x that satisfies this equation is x = 3. Therefore, the correct answer is.