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Math Help - Further Probability

  1. #1
    Member classicstrings's Avatar
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    Further Probability

    If no digit can be used more than once, find how many numbers can be formed from the digits 3,4,5,6,7,8 which are six-digit numbers and even.
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    Grand Panjandrum
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    Quote Originally Posted by classicstrings View Post
    If no digit can be used more than once, find how many numbers can be formed from the digits 3,4,5,6,7,8 which are six-digit numbers and even.
    Since you have six digits and are asked for six digit numbers, this is
    equivalent to asking how many permutations are there of six distinct
    objects.

    The number of permutations of six distinct objects is 6!=720.

    You can see that this is the case as the first digit in the perm can be
    any of the six, the second any of the remaining 5, the third any of the
    remaining 4, ... The total number of permutations is the product of
    these, that is 6x5x4x3x2x1=6!

    RonL
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  3. #3
    Member classicstrings's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Since you have six digits and are asked for six digit numbers, this is
    equivalent to asking how many permutations are there of six distinct
    objects.

    The number of permutations of six distinct objects is 6!=720.

    You can see that this is the case as the first digit in the perm can be
    any of the six, the second any of the remaining 5, the third any of the
    remaining 4, ... The total number of permutations is the product of
    these, that is 6x5x4x3x2x1=6!

    RonL
    I realise that but what if they have to be even?

    SO the last digit has to be 2,4,6

    So, 6x5x3x2x1x1 correct?
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    Quote Originally Posted by classicstrings View Post
    I realise that but what if they have to be even?

    SO the last digit has to be 2,4,6

    So, 6x5x3x2x1x1 correct?
    Oops, sory about that, I did not read the question carefully enough


    For the numbers to be even work from the least to the most significant
    digit.

    The least significant can be any of 3, then the next any of the remaining 5,
    and so on as before.

    So then we have 3x5x4x3x2x1=360

    RonL
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  5. #5
    Member classicstrings's Avatar
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    Can you tell me why my logic is incorrect?

    I think about it this way, at the start you have 6 digits so it can be any of the six.

    Then you are left with 5 you can still choose any of the five.

    Then if the firs two were both even you cant choose the last even so the possibilities are now 3.

    Then it's 2, then 1, and last you are left with 1 even number.

    So 6x5x3x2x1x1. But it's wrong
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    Quote Originally Posted by classicstrings View Post
    Can you tell me why my logic is incorrect?

    I think about it this way, at the start you have 6 digits so it can be any of the six.

    Then you are left with 5 you can still choose any of the five.

    Then if the firs two were both even you cant choose the last even so the possibilities are now 3.

    Then it's 2, then 1, and last you are left with 1 even number.

    So 6x5x3x2x1x1. But it's wrong
    You don't really have the freedom to choose any of the six digits for
    the most significant, (and 5 for the next ..), because you must have
    an even digit left when you reach the least significant digit. But if you
    have been selecting the earlier digits from all of the available digits
    you may have no even digits left at the end.

    The number of choices later in the sequence available if you place no
    constraint on how you choose the first two depend on what you actually
    chose for the first two so multiplying the numbers together this way does not work.

    Much better to start choosing from the other end, where the constraint
    can be dealt with easily.

    RonL
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  7. #7
    Member classicstrings's Avatar
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    Thanks for the explanation, from now on i'll always eliminate the constraint first, then the rest follows. Cheers.
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