WAEC Further Mathematics 2024 Paper

Concept:Since repetition is not allowed, each selection of different digits must be arranged in order, so we use permutations.Explanation:There are 7 available digits: $2,3,4,5,6,7,8$. To form a three-digit number, we select and arrange 3 digits out of these 7 digits. The number of such ordered arrangements is given by $^{7}P_{3}$. $^{7}P_{3} = \frac{7!}{(7-3)!} = \frac{7!}{4!}$. This can be written as $\frac{7 \times 6 \times 5 \times 4!}{4!}$. Cancelling $4!$ gives $7 \times 6 \times 5$. So the total number of three-digit numbers is $7 \times 6 \times 5 = 210$. Alternatively, the first digit has 7 choices, the second digit has 6 remaining choices, and the third digit has 5 remaining choices. Thus, total $= 7 \times 6 \times 5 = 210$.Answer:$210$, which is Option B.