The number of seven digit numbers divisible by 9 formed with the digits ,1,2,3,4,5,6,7,8,9 without repetition is (A) 7! (B) ^9P_7` (C) `3(7!)` (D) `4(7!)`

Options

A. `2//9`
B. `1//5`
C. `1//3`
D. `1//9`

Answer

To find the number of seven-digit numbers divisible by 9 formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 without repetition, we can follow these steps:

### Step 1: Understanding the divisibility rule for 9
A number is divisible by 9 if the sum of its digits is divisible by 9.

### Step 2: Calculate the total sum of digits
The digits available are 1, 2, 3, 4, 5, 6, 7, 8, and 9. The total sum of these digits is:
\[ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \]
Since 45 is divisible by 9, any combination of 7 digits from these 9 digits will also be divisible by 9 if the sum of the remaining 2 digits is divisible by 9.

### Step 3: Identify pairs of digits whose sum is divisible by 9
We need to find pairs of digits that can be removed such that their sum is divisible by 9. The pairs are:
- (1, 8) → 1 + 8 = 9
- (2, 7) → 2 + 7 = 9
- (3, 6) → 3 + 6 = 9
- (4, 5) → 4 + 5 = 9

Thus, we have 4 valid pairs.

### Step 4: Calculate the number of ways to form 7-digit numbers
For each pair of digits that we can remove, we are left with 7 digits. The number of ways to arrange 7 digits is given by \(7!\).

### Step 5: Total arrangements
Since there are 4 pairs of digits that can be removed, the total number of 7-digit numbers that can be formed is:
\[ \text{Total} = 4 \times 7! \]

### Conclusion
Thus, the number of seven-digit numbers divisible by 9 formed with the digits 1 to 9 without repetition is:
\[ \text{Total} = 4(7!) \]

### Final Answer
The correct option is (D) \(4(7!)\).

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