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Math Help - number of even integers with no repeated digits between 20000 and 70000

  1. #1
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    number of even integers with no repeated digits between 20000 and 70000

    I have been given the problem to solve how many even integers with no repeated digits between 20,000 and 70,000.

    my first thought was to have the set X = {0,1,2,3,4,5,6,7,8,9}

    and then propose there are 5 positions in which these numbers can be placed, being ABCDE

    therefore E has to be either {0,2,4,6,8} => |E| = 5

    D is any other of the 10 ten numbers excluding E => |D| = 8 and so on.

    therefore i originally thought that the answer would be 5x9x8x7x2 (using the mulitplication prinicple)

    As A is either 2,3,4,5,6,7 and cant be what B,C,D was therefore |A| = 2, i know that this is not true as BCD can be numbers outside of this region. but i am unclear as to how to find |A| without knowing what values B,C,D took on? any help would be appreciated.

    Thanks
    Reece
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  2. #2
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    Re: number of even integers with no repeated digits between 20000 and 70000

    First digit even 3 ways, last digit even 4 ways, other digits 8*7*6 ways.

    First digit odd 3 ways, last digit even 5 ways, other digits 8*7*6 ways.
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  3. #3
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    Re: number of even integers with no repeated digits between 20000 and 70000

    Hello, myenemy!

    I got a different answer . . .


    How many even integers with no repeated digits between 20,000 and 70,000?

    We have a five-digit number:. ABCDE

    The first digit A can be: \{2,3,4,5,6\} ... 5 choices.

    The last digit E can be: \{0,2,4,6,8\} ... 5 choices.


    There are two cases to consider.


    [2] E = 0 or 8 . . . 2 choices.

    . . Then there are 5 choices for A.

    . . The other 3 digits can be chosen in _8P_3 = 336 ways.

    Hence, there are: 2\cdot5\cdot336 \,=\,3360 numbers that end in 0 or 8.


    [2] E = 2, 4 or 6 . . . 3 choices.

    . . Then there are only 4 choices for A.

    . . The other 3 digits can be chosen in _8P_3 = 336 ways.

    Hence, there are: 3\cdot4\cdot336 \,=\,4032 numbers that end in 2, 4 or 6.


    Therefore, there are: 3360 + 4032 \:=\:7392 even numbers
    . . between 20,000 and 70,000 with no repeated digits.
    Thanks from a tutor
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  4. #4
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    Re: number of even integers with no repeated digits between 20000 and 70000

    Thanks Soroban. I made a silly little mistake. It can't start with a 7.


    Quote Originally Posted by a tutor View Post
    First digit even 3 ways, last digit even 4 ways, other digits 8*7*6 ways.

    First digit odd 3 ways, last digit even 5 ways, other digits 8*7*6 ways.
    First digit even 3 ways, last digit even 4 ways, other digits 8*7*6 ways.

    First digit odd 2 ways, last digit even 5 ways, other digits 8*7*6 ways.

    So 3.4.8.7.6 + 2.5.8.7.6 = 7392 in agreement with Soroban.
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