1. **State the problem:** We need to find the number of 6-character passwords formed from letters {b, f, g, k, m}, numbers {3, 5, 7, 9}, and symbols {*, !, @} with given restrictions. 2. **Count total characters and available sets:** - Letters = 5 - Numbers = 4 - Symbols = 3 - Total characters = 5 + 4 + 3 = 12 - Each character used at most once --- (a) **No restrictions:** - Number of ways to choose and arrange any 6 characters out of 12 different characters without repetition is a permutation: $$ P(12,6) = \frac{12!}{(12-6)!} = \frac{12!}{6!} $$ Calculate: $$ 12 \times 11 \times 10 \times 9 \times 8 \times 7 = 665280 $$ --- (b) **Password starts and ends with a symbol:** - Positions: 1 and 6 must be symbols. - Choose symbol for position 1: 3 choices - Choose symbol for position 6: since no repetition, 2 choices - Remaining 4 positions (positions 2,3,4,5) chosen from remaining characters: Remaining characters after choosing 2 symbols: $12 - 2 = 10$ Number of ways to choose and arrange 4 characters from 10: $$ P(10,4) = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040 $$ Total passwords: $$ 3 \times 2 \times 5040 = 30240 $$ --- (c) **Password conditions:** - Starts with (symbol then number) OR (number then symbol) - Ends with 2 letters Step 1: Count ways for the first two characters: - Symbol then number: $$ 3 \text{(symbols)} \times 4 \text{(numbers)} = 12 $$ - Number then symbol: $$ 4 \times 3 = 12 $$ Total for first two characters: $$ 12 + 12 = 24 $$ Step 2: Last two characters are letters: - Must be 2 letters chosen and arranged from 5 letters without repetition: $$ P(5,2) = 5 \times 4 = 20 $$ Step 3: Middle two characters (positions 3 and 4): - Characters remaining after first 2 and last 2 chosen: Total characters: 12 Used in first 2 + last 2: 4 Remaining: 12 - 4 = 8 Number of ways to choose and arrange 2 characters from 8: $$ P(8,2) = 8 \times 7 = 56 $$ Total passwords: $$ 24 \times 56 \times 20 = 26880 $$