How many 3 digit numbers can be formed if the leading and ending digit cannot be zero

Since we can't use a zero at the start and end, then we have 810 different ways.

If we can't have a leading digit of zero, then we still have 9 available digits to choose from (9P1)
Now, the second term can be of any number, so we will have 10 different digits to choose 1 (10P1)
Now, the final term cannot be a zero, so we will have 9 available digits to choose from (9P1)

Hence, our final number of ways is: 9 × 10 × 9 = 810 ways.

(A) The Leading Digit Cannot Be Zero 
9 Choices For First Digit (1, 2, ..., 9) 
10 Choices For The Second Digit 
10 Choices For The Third Digit 
9*10*10 = 900 

(B) The Leading Digit Cannot Be Zero And No Repetition Of Digits Is Allowed. ( Im just going to write lower case letters now xD)
9 choices for first digit (1, 2, ..., 9) 
9 choices for the second digit, anything different from the first 
8 choices for the third digit, anything different from first and second 
9*9*8 = 648 

(C) The leading digit cannot be zero and the number is a multiple of 5 
9 choices for first digit (1, 2, ..., 9) 
10 choices for the second digit 
2 choices for the third digit (0 or 5) 
9*10*2 = 180 

(D) the number is at least 400 
6 choices for first digit (4, 5, ..., 9) 
10 choices for the second digit 
10 choices for the third digit 
6*10*10 = 600

Ithink the answer is a