---

Solution

---

Solution

---

Solution

---

Solution

---

Solution

Letters: A, A, G, I, N Words starting with A: $ \frac{4!}{2!}=12 $ Starting with G: $ \frac{4!}{2!}=12 $ Starting with I: $ \frac{4!}{2!}=12 $ Total till I: $12+12+12=36$ Next start with N Remaining: A, A, G, I Arrangements: $ \frac{4!}{2!}=12 $ So positions $37$ to $48$ $49^{th}$ word = NAAGI $50^{th}$ word = NAAIG

---

Solution

Given, $^nP_4 = 20 \times ^nP_2$$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$$\frac{1}{(n - 4)!} = \frac{20}{(n - 2)(n - 3)(n - 4)!}$$(n - 2)(n - 3) = 20$$n^2 - 3n - 2n + 6 = 20$$n^2 - 5n + 6 - 20 = 0$$n^2 - 5n - 14 = 0$$(n - 7)(n + 2) = 0$This gives $n = 7$ or $n = -2$.Since $n$ must be a positive integer ($n \geq 4$ for $^nP_4$), $n = -2$ is not possible.$\therefore n = 7$

---

Solution

$\frac{^nP_5}{^nP_3} = \frac{n!/(n-5)!}{n!/(n-3)!}$ $= \frac{(n-3)!}{(n-5)!}$ $= (n-3)(n-4)$ So, $(n-3)(n-4) = 20$ $n^2 - 7n + 12 = 20$ $n^2 - 7n - 8 = 0$ $(n-8)(n+1) = 0$ $n = 8$