## Task B-3.4.

How many nine-digit numbers divisible by 75 are there, all of whose digits are different, and the hundreds digit is 7?
## Solution.

The desired number is divisible by 75, so it is also divisible by 25. Since it is divisible by 25 and has all distinct digits, the last three digits of this number are either 725 or 750.

Let's determine the first 6 digits as well.

The desired number must be divisible by 3. Since the sum of all 10 possible digits $0+1+2+\ldots+8+9=45$, we can remove one of the digits $0,3,6$ or 9 to make the number divisible by 3.

If we remove 0, the last three digits must be 725, and the other 6 can be chosen in

$6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720$ ways.

If we remove 3, 6, or 9, and the last three digits are 750, the other 6 can be chosen in $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720$ ways, which is a total of $3 \cdot 720=2160$ ways.

If we remove 3, 6, or 9, and the last three digits are 725, the first digit cannot be 0, so we choose them in

$5 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=600$ ways, which is a total of $3 \cdot 600=1800$ ways.

Therefore, the total number of desired nine-digit numbers is $720+2160+1800=4680$.