Given: \(\frac{^6C_r}{^6P_r} = \frac{1}{6}\) Recall the formulas: \(^nC_r = \frac{n!}{r!(n-r)!}, \quad ^nP_r = \frac{n!}{r!}\) Substitute \( n = 6 \): \(\frac{^6C_r}{^6P_r} = \frac{\dfrac{6!}{r!(6-r)!}}{\dfrac{6!}{r!}} = \frac{6!}{r!(6-r)!} \times \frac{r!}{6!} = \frac{1}{(6-r)!}\) So the equation simplifies to: \(\frac{1}{(6-r)!} = \frac{1}{6}\) Therefore: (6-r)! = 6 We know that 3! = 6, so: 6 - r = 3 \(\implies\) r = 3