Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition). the total number of such numbers that can be formed isChoices:- 216 214 212 218

To solve this problem, we need to understand the divisibility rule of 9. A number is divisible by 9 if the sum of its digits is divisible by 9.

Step 1: Sum of the digits 0, 1, 2, 3, 4, 7, 8 is 25. The closest multiple of 9 less than 25 is 18. So, we need to exclude digits that sum up to 7 (25-18) to make the sum of the digits of the number divisible by 9. The combinations that sum up to 7 are (0,7), (1,2,4), and (3,4).

Step 2: We have 5 digits left to form the number. The first digit can be any of the 4 non-zero digits, the second digit can be any of the remaining 4 digits (including 0), the third digit can be any of the remaining 3 digits, the fourth digit can be any of the remaining 2 digits, and the last digit is the remaining digit.

Step 3: So, the total number of such numbers is 44321 = 96.

Step 4: We repeat the same process for each combination that sums up to 7. So, the total number of such numbers is 96*3 = 288.

However, we have over-counted the numbers that start with 0, which are not 5-digit numbers.

Step 5: For each combination that sums up to 7, there are 321 = 6 such numbers that start with 0. So, the total number of such numbers is 6*3 = 18.

Step 6: Subtract the over-counted numbers from the total number of numbers. So, the final answer is 288 - 18 = 270.

However, none of the given choices match this answer. There might be a mistake in the problem or the choices.