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28 February, 21:58

Find the number of five-digit multiples of 5, where all the digits are different, and the second digit (from the left) is odd.

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  1. 28 February, 22:02
    0
    Given:

    candidates for digits from left:

    1. [1-9] (leftmost, 5 digit-number)

    2. [13579] odd

    3. [0-9]

    4. [0-9]

    5. [05] (multiple of 5).

    We will start with digits with most constraints (5,2,1,3,4)

    For the fifth digit, number ends with either 0 or 5, so two cases.

    Case 1:

    5th digit is a 0 (one choice)

    2nd digit can be chosen from [1,3,5,7,9] so 5 choices

    1st digit can be chosen from [1-9] less second digit, so 8 choices

    3rd digit can be chosen from remaining 7 choices

    4th digit can be chosen from remaining 6 choices

    for a total of 1*5*8*7*6=1680 arrangements

    Case 2

    5th digit is a 5 (one choice)

    2nd digit can be chosen from [1379], so 4 choices

    1st digit can be chosen from [1-4,6-9] less 2nd digit, so 8-1=7 choices

    3rd digit can be chosen from remaining 7 choices

    4th digit can be chosen from remaining 6 choices

    for a total of 1*4*7*7*6=1176

    Case 1 + Case 2

    = total number of arrangements for 5 digit multiples of 5 with second digit odd

    = 1680+1176

    = 2856
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