1. **Stating the problem:** We want to find the number of words formed using the letters of the word DEPARTMENT with each letter used at most once. 2. The word DEPARTMENT has 10 letters: D, E, P, A, R, T, M, E, N, T. 3. Letters E and T are repeated twice. So the letters are D, E, E, P, A, R, T, T, M, N. ### Part i) Using all letters: 4. We want to find the number of distinct permutations of all 10 letters, accounting for duplicates. 5. The total number of permutations of 10 letters with E repeated twice and T repeated twice is: $$\frac{10!}{2! \times 2!}$$ 6. Calculate factorial values: $$10! = 3628800$$ $$2! = 2$$ 7. Therefore, $$\frac{3628800}{2 \times 2} = \frac{3628800}{4} = 907200$$ So, 907200 distinct words can be formed using all letters. ### Part ii) Some or all letters may be omitted: 8. Here, any subset of letters can be used to form words, with each letter used at most once. 9. For each length $k$ from 1 to 10, calculate the number of distinct arrangements of $k$ letters chosen from DEPARTMENT, considering repetition. 10. The total number is the sum over all $k$ of the number of distinct permutations of length $k$. 11. This is equivalent to counting all distinct permutations of all subsets of letters from the multiset \{D, E, E, P, A, R, T, T, M, N\}. 12. The formula for the total number of distinct words is: $$\sum_{k=1}^{10} \text{number of distinct permutations of length } k$$ 13. For such multisets, the total number of distinct words that can be formed (using at most the available letters) is: $$\prod \left(1 + n_i\right) - 1$$ where $n_i$ is the multiplicity of each distinct letter, because for each letter we can use 0 up to $n_i$ copies, except we exclude the empty word. 14. Here, letters and their multiplicities: - D:1 - E:2 - P:1 - A:1 - R:1 - T:2 - M:1 - N:1 15. So, $$\left(1+1\right)\times \left(1+2\right) \times \left(1+1\right) \times \left(1+1\right) \times \left(1+1\right) \times \left(1+2\right) \times \left(1+1\right) \times \left(1+1\right) -1$$ $$= 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2 -1$$ 16. Calculate the product stepwise: $$2 \times 3 = 6$$ $$6 \times 2 = 12$$ $$12 \times 2 = 24$$ $$24 \times 2 = 48$$ $$48 \times 3 = 144$$ $$144 \times 2 = 288$$ $$288 \times 2 = 576$$ 17. Subtract 1 for the empty word: $$576 - 1 = 575$$ So, 575 distinct words can be formed using some or all letters from DEPARTMENT. **Final Answers:** - i) Using all letters: 907200 - ii) Using some or all letters: 575