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    Using the numbers 0, 1, 2, ..., 9 as digits, how many four-digit numbers exist for which the following three conditions hold simultaneously: (1) all digits are different, (2) two digits are even numbers, and (3) two digits are odd numbers? Recall that 0 is an even number, and that a four-digit number by definition does not start by 0.

    finding it difficult to process how to combine all the elements of the question. any help would be appreciated
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    (Original post by marmslboro)
    Using the numbers 0, 1, 2, ..., 9 as digits, how many four-digit numbers exist for which the following three conditions hold simultaneously: (1) all digits are different, (2) two digits are even numbers, and (3) two digits are odd numbers? Recall that 0 is an even number, and that a four-digit number by definition does not start by 0.

    finding it difficult to process how to combine all the elements of the question. any help would be appreciated
    Gonna go ahead and make an educated guess that
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    You're a 1st year from Lboro and this question is part of your coursework on Introduction to Probability and Statistics.


    Anyway, been there and done this question last year (it's recycled it seems!) so you can approach it from the following angle:

    Say that the 4 digit number can be written in the form (a,b,c,d) such that a \neq b \neq c \neq d and a \neq 0 due to conditions.
    How many possible options are there for a? How many of those are odd/even?

    From this you can deduce 6 different forms in which the number can be represented (representing each digits as even or odd). Write them out.
    Then consider how many combinations of each form there are before adding them up.
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    (Original post by marmslboro)
    Using the numbers 0, 1, 2, ..., 9 as digits, how many four-digit numbers exist for which the following three conditions hold simultaneously: (1) all digits are different, (2) two digits are even numbers, and (3) two digits are odd numbers? Recall that 0 is an even number, and that a four-digit number by definition does not start by 0.

    finding it difficult to process how to combine all the elements of the question. any help would be appreciated
    This is more of a combinatorics question than a probability question, isn't it?

    Consider any digit (other than zero) as the first digit; there are 9 possible choices at this stage. You then have to choose a different digit [condition 1], but zero can be used now - so there are again 9 choices. For the third digit, you have 8 choices remaining, and then 7 choices for the fourth and final digit.

    How many four-digit numbers with all distinct digits does this give us in total?

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    9*9*8*7 = 4,536 such numbers.

    Try working from here; play around with the other two conditions, and see what you can come-up with...

    I hope that this helps!
 
 
 
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