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28 December, 01:50

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

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  1. 28 December, 02:15
    0
    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.
  2. 28 December, 02:16
    0
    (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
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