1. **Problem Statement:** We have 6 characters to form a password from letters \(\{b,f,g,k,m\}\), numbers \(\{3,5,7,9\}\), and symbols \(\{*,!,@\}\), with each character used at most once. **(a)** Number of 6-character passwords with no restrictions. 2. Total characters available: \(5 + 4 + 3 = 12\). 3. Since repetition is not allowed, the number of 6-character passwords is the number of 6-permutations from 12 characters: $$ P(12,6) = \frac{12!}{(12-6)!} = \frac{12!}{6!} $$ 4. Calculate: $$12! = 479001600, \quad 6! = 720,$$ $$P(12,6) = \frac{479001600}{720} = 665280.$$ **(b)** Password starts and ends with a symbol. 5. Positions 1 and 6 must be symbols, and symbols are \(\{*,!,@\}\), 3 in total. 6. Choose symbol for position 1: 3 choices. 7. Choose symbol for position 6 (distinct from position 1): 2 choices. 8. Remaining 4 positions are to be filled from the remaining 12 - 2 = 10 characters (since 2 symbols used). 9. Number of ways to arrange 4 characters from 10 without repetition: $$P(10,4) = \frac{10!}{6!} = 5040.$$ 10. Total number of such passwords: $$3 \times 2 \times 5040 = 30240.$$ **(c)** Password starts with either (symbol then number) or (number then symbol), and ends with 2 letters. 11. For the last two letters: - Letters available: 5. - Number of ways to choose and order 2 letters: \(P(5,2) = \frac{5!}{3!} = 20\). 12. For the first two positions: - Case 1: First is symbol (3 choices), second is number (4 choices), total \(3 \times 4 = 12\). - Case 2: First is number (4 choices), second is symbol (3 choices), total \(4 \times 3 = 12\). - Total for first two positions: \(12 + 12 = 24\). 13. Number of characters used so far: 2 (first two positions) + 2 (last two letters) = 4. 14. Remaining characters: 12 - 4 = 8. 15. Positions 3,4,5 are to be filled from these 8 characters without repetition: $$P(8,3) = \frac{8!}{5!} = 336.$$ 16. Total number of such passwords: $$24 \times 336 \times 20 = 161280.$$