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

(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

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.