Five digit numbers divisible by 9 are to be formed by using the digits 0 1 2 3 4 7 8 The total number of such numbers is equal to

Options

A. 216
B. 214
C. 212
D. 200

Answer

To solve the problem of finding the total number of five-digit numbers divisible by 9 that can be formed using the digits 0, 1, 2, 3, 4, 7, and 8, we will follow these steps:

### Step 1: Calculate the sum of the given digits
The first step is to find the sum of all the digits provided:
\[ 0 + 1 + 2 + 3 + 4 + 7 + 8 = 25 \]

### Step 2: Determine the conditions for divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, we need to find combinations of 5 digits from the 7 digits such that their sum is either 9 or 18 (since the maximum sum of 5 digits can be 25).

### Step 3: Check for the sum of 9
To achieve a sum of 9, we need to select 5 digits whose total is 9. However, the smallest sum of any 5 digits from the set {0, 1, 2, 3, 4, 7, 8} is:
\[ 0 + 1 + 2 + 3 + 4 = 10 \]
Thus, it is impossible to form a combination of 5 digits that sums to 9. Therefore, we discard this case.

### Step 4: Check for the sum of 18
Next, we check for combinations that sum to 18. Since the total sum of all digits is 25, we need to exclude digits that sum to:
\[ 25 - 18 = 7 \]
We can achieve this by excluding the following pairs of digits:
1. Exclude 0 and 7: Remaining digits are {1, 2, 3, 4, 8} (sum = 18)
2. Exclude 3 and 4: Remaining digits are {0, 1, 2, 7, 8} (sum = 18)

### Step 5: Calculate the number of valid combinations for each case
#### Case 1: Excluding 0 and 7
The remaining digits are {1, 2, 3, 4, 8}. All 5 digits can be arranged in:
\[ 5! = 120 \]

#### Case 2: Excluding 3 and 4
The remaining digits are {0, 1, 2, 7, 8}. However, since a five-digit number cannot start with 0, we need to fix 0 in a non-leading position. We can fix 0 in one of the last four positions and arrange the other four digits (1, 2, 7, 8) in the first position:
- Choose one of the first 4 positions for 0 (4 choices).
- Arrange the remaining 4 digits in the other positions (4! arrangements):
\[ 4 \times 4! = 4 \times 24 = 96 \]

### Step 6: Total the valid combinations
Now, we add the valid combinations from both cases:
\[ 120 + 96 = 216 \]

### Conclusion
Thus, the total number of five-digit numbers divisible by 9 that can be formed using the digits 0, 1, 2, 3, 4, 7, and 8 is:
\[ \boxed{216} \]