(i) We have, $\\,^{5}P_r$ = 2 $\\,^{6}P_{r-1}$ ⇒ $\frac{5!}{(5-r)!}$ = 2 × $\frac{6!}{(6-r+1)!}$ ⇒ $\frac{5!}{(5-r)!}$ = $\frac{2\cdot6\cdot5!}{(7-r)!}$ ⇒ 12 (5 – r) ! = (7 – r)! ⇒ 12 (5 – r)! = (7 – r) (6 – r) (5 – r)! ⇒ $r^{2}$ – 13r + 30 = 0 ⇒ (r – 3) (r – 10) = 0 ⇒ r = 3, 10 But r ≠ 10 (Since r ≤ 5) ∴ r = 3. (ii) $\\,^{5}P_r$ = $\\,^{6}P_{r-1}$ ⇒ $\frac{5!}{(5-r)!}$ = $\frac{6!}{(7-r)!}$ ⇒ $\frac{1}{(5-r)!}$ = $\frac{6}{(7-r)!}$ ⇒ (7 – r) (6 – r) (5 – r)! = 6(5 – r)! ⇒ $r^{2}$ – 13r + 36 = 0 ⇒ (r – 4)(r – 9) = 0 ⇒ r = 4, 9 but r ≠ 9 (Since r ≤ 5) ∴ r = 4.