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Math Help - Counting

  1. #16
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    Quote Originally Posted by jvignacio View Post
    wow thats heaps confusing to understand... Is there another way of doing this? I forgot to consider the numbers greater than 3 can have a zero on the second digit
    There's no easy way to do these, it's really just using logic:

    Numbers that begin with 1:
    1x?x?x?x?x? (first digit MUST be 1, so 1 possibility)
    1x?x?x?x?x5 (last digit must be 0,2,4,6,or 8 so 5 possibilities)
    1x8x7x6x5x5 (8 possibilites left for the second digit and so on)

    Numbers that begin with 2:
    1x?x?x?x?x? (first digit MUST be 2, so 1 possibility)
    1x?x?x?x?x4 (last digit must be 0,4,6,or 8 (can't be 2) so 4 possibilities)
    1x8x7x6x5x4 (8 possibilites left for the second digit and so on)

    Numbers that begin with 30:
    1x1x?x?x?x? (Must be 3 in first and 0 in second)
    1x1x?x?x?x4 (To be even, last digit must be 2,4,6,8 as 0 is already used)
    1x1x7x6x5x4 ( 7 possibilities left for the third digit and so on)

    Adding these underlined values gives the total number of 6-digit even numbers (no repeating digits) less than 310 000.

    So take this total away from (c) {the corrected version} to get the number of required numbers greater than (or equal to) 310 000. If you want the number greater tha 310 000, you'll have to subtract 1 (to exclude 310 000 itself from being counted).

    Makes sense to me. Hope it makes sense to you too!
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  2. #17
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    Quote Originally Posted by Debsta View Post
    There's no easy way to do these, it's really just using logic:

    Numbers that begin with 1:
    1x?x?x?x?x? (first digit MUST be 1, so 1 possibility)
    1x?x?x?x?x5 (last digit must be 0,2,4,6,or 8 so 5 possibilities)
    1x8x7x6x5x5 (8 possibilites left for the second digit and so on)

    Numbers that begin with 2:
    1x?x?x?x?x? (first digit MUST be 2, so 1 possibility)
    1x?x?x?x?x4 (last digit must be 0,4,6,or 8 (can't be 2) so 4 possibilities)
    1x8x7x6x5x4 (8 possibilites left for the second digit and so on)

    Numbers that begin with 30:
    1x1x?x?x?x? (Must be 3 in first and 0 in second)
    1x1x?x?x?x4 (To be even, last digit must be 2,4,6,8 as 0 is already used)
    1x1x7x6x5x4 ( 7 possibilities left for the third digit and so on)

    Adding these underlined values gives the total number of 6-digit even numbers (no repeating digits) less than 310 000.

    So take this total away from (c) {the corrected version} to get the number of required numbers greater than (or equal to) 310 000. If you want the number greater tha 310 000, you'll have to subtract 1 (to exclude 310 000 itself from being counted).

    Makes sense to me. Hope it makes sense to you too!
    Yes makes sense now thanks heaps ay for helping me.... really appreaciate the help.
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  3. #18
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    Quote Originally Posted by jvignacio View Post
    Yes makes sense now thanks heaps ay for helping me.... really appreaciate the help.
    It's been a pleasure - love these sort of problems. I'm off now to get a life!!
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