1. **Problem statement:** (a) Find how many 3-digit and 4-digit numbers can be formed using digits 4, 5, 6, 7, 8 without repetition. (b) Find how many 3-digit numbers are (i) greater than 600 and (ii) less than 600. 2. **Formula and rules:** - Number of permutations of $n$ distinct digits taken $r$ at a time is given by $$P(n,r) = \frac{n!}{(n-r)!}$$ - Digits cannot repeat. - For 3-digit numbers, the first digit cannot be zero (not applicable here since digits are 4,5,6,7,8). 3. **Part (a) 3-digit numbers:** - Total digits = 5 - Number of digits to choose = 3 - Number of 3-digit numbers = $$P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60$$ 4. **Part (a) 4-digit numbers:** - Number of digits to choose = 4 - Number of 4-digit numbers = $$P(5,4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 120$$ 5. **Part (b)(i) 3-digit numbers greater than 600:** - First digit must be 6, 7, or 8 (3 choices) - Remaining 2 digits chosen from remaining 4 digits without repetition - Number of ways to choose last 2 digits = $$P(4,2) = \frac{4!}{2!} = \frac{24}{2} = 12$$ - Total numbers = $$3 \times 12 = 36$$ 6. **Part (b)(ii) 3-digit numbers less than 600:** - First digit must be 4 or 5 (2 choices) - Remaining 2 digits chosen from remaining 4 digits without repetition - Number of ways to choose last 2 digits = $$P(4,2) = 12$$ - Total numbers = $$2 \times 12 = 24$$ **Final answers:** - (a) 3-digit numbers: 60 - (a) 4-digit numbers: 120 - (b)(i) 3-digit numbers > 600: 36 - (b)(ii) 3-digit numbers < 600: 24