1. **Problem:** Find the number of arrangements of the letters of the given words. 2. **Formula:** The number of arrangements of $n$ letters where some letters repeat is given by $$\frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}$$ where $n$ is the total number of letters and $n_1, n_2, \ldots, n_k$ are the frequencies of the repeated letters. 3. **Calculations:** - (i) PAKPATTAN: Letters = P(2), A(3), K(1), T(2), N(1) Total letters $n=9$ Number of arrangements = $$\frac{9!}{2! \times 3! \times 2! \times 1! \times 1!} = \frac{362880}{2 \times 6 \times 2} = \frac{362880}{24} = 15120$$ - (ii) PAKISTAN: Letters = P(1), A(2), K(1), I(1), S(1), T(1), N(1) Total letters $n=8$ Number of arrangements = $$\frac{8!}{2!} = \frac{40320}{2} = 20160$$ - (iii) MATHEMATICS: Letters = M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1) Total letters $n=11$ Number of arrangements = $$\frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{8} = 4989600$$ - (iv) ASSASSINATION: Letters = A(3), S(4), I(2), N(2), T(1), O(1) Total letters $n=12$ Number of arrangements = $$\frac{12!}{3! \times 4! \times 2! \times 2!} = \frac{479001600}{6 \times 24 \times 2 \times 2} = \frac{479001600}{576} = 831600$$ **Final answers:** (i) 15120 (ii) 20160 (iii) 4989600 (iv) 831600