Final answer:
To form an eight-digit number divisible by 9 with non-repeating digits, you must create a set of 8 digits whose sum is a multiple of 9. Considering all arrangements, there are 5 sets of such digits, including one set with digits 1 to 8 and four sets that include the digit 9 and omit one of the digits 1 to 4. For each set, there are 7! ways to arrange the digits, leading to the answer of 40 × 7!.
Step-by-step explanation:
To find out the number of ways an eight-digit number divisible by 9 can be formed with non-repeating digits from 0 to 9, we need to consider the properties of numbers divisible by 9. A number is divisible by 9 if the sum of its digits is a multiple of 9. With this in mind, we need to find sets of 8 different digits such that their sum is 9, 18, 27, and so on up to the maximum possible which is 44 (using the digits 1 to 8). As 0 cannot contribute to the sum for divisibility, it can be placed in any of the remaining 7 positions after the first, giving 7 choices for its position.
Once we have fixed the position for the digit 0, the remaining 7 digits can be arranged in 7! ways. If the digits 1 to 8 are used, their sum is 36, which is a multiple of 9. Thus this set of 8 digits (including '0' at a chosen position) can be arranged in 7! ways. However, if 9 is included in the set, we must exclude one of the other digits to maintain the sum as a multiple of 9. There are four possible sets that include 9 and omit one digit from 1 to 8: omitting 1 (sum is 44), 2 (sum is 43), 3 (sum is 42), or 4 (sum is 41) - all these sums are multiples of 9.
For each of these sets, with 9 included, there are still 7! arrangements, once '0' is placed in one of 7 spots. Therefore, we have one arrangement corresponding to the set with 1 to 8, plus four possible sets when 9 is included, each with 7! arrangements, totalling 5 * 7! possibilities. Hence, the number of ways an eight-digit number divisible by 9 can be formed is 5 × 7!, which is option C. 40 (7!).