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Math Help - Counting

  1. #1
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    Counting

    Hi there I have a question...

    Can anyone explain me this question which is along with its answer!

    Count the set S of 3 digit numbers which begin or end
    with an even digit.
    Assume that 0 is even but a number cannot begin with 0.
    The set is the union of the two subsets:
    The set B of three digit numbers that begin
    with 2, 4, 6 or 8.
    This set has cardinality
    (4)(10)(10).
    (why?)

    The set C of three digit numbers that end with
    0, 2, 4, 6, or 8 and do not begin with 0.
    This set has cardinality
    (5)(9)(10). <<How come here it is (5)(9)(10) instead of (5)(10)(10)?
    (why?)
    Now we use the inclusion-exclusion principle
    to eliminate the overlap of sets B and C.
    Their intersection:
    The 3 digit numbers that begin with 2, 4, 6, or 8 and end
    with 0, 2, 4, 6, or 8.
    The intersection has the cardinality
    (4)(10)(5)
    Hence the cardinality is
    400 + 450 - 200 = 650.
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  2. #2
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    It's 5*9*10 because there are 900 three digit numbers between 100 and 999 and half of these are even, therefore 900/2 = 450.
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  3. #3
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    The first set has a cardinality of 400 because...

    There are 4 different numbers which can be the 10^2 digit (2, 4, 6, 8)

    10 numbers which can be the digit in the tens column (0-9)

    And 10 numbers which can be the digit in the units column (0-9)
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  4. #4
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    Quote Originally Posted by sean.1986 View Post
    It's 5*9*10 because there are 900 three digit numbers between 100 and 999 and half of these are even, therefore 900/2 = 450.
    I still don't get it How come its 5*9*10? shouldn't it be 9*10*5? where 5 is units 10 is tens and 9 is hundreds?
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  5. #5
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    Quote Originally Posted by robocop_911 View Post
    I still don't get it How come its 5*9*10? shouldn't it be 9*10*5? where 5 is units 10 is tens and 9 is hundreds?
    Basically, yeah. The commutative laws of multiplication mean they're the same.
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  6. #6
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    Hello, robocop_911!

    Count the set S of 3-digit numbers which begin or end with an even digit.

    Assume that 0 is even but a number cannot begin with 0.

    The set is the union of the two subsets:

    The set B of 3-digit numbers that begin with 2, 4, 6 or 8.
    The first digit can be any of {2,4,6,8}: .4 choices.
    The second digit can be any of the 10 digits.
    The third digit can be any of the 10 digits.

    Hence, there are: . (4)(10)(10) \:=\:400 numbers in set B \;\;\hdots\;\; n(B) = 400



    The set C of 3-digit numbers that end with {0,2,4,6,8} and do not begin with 0.
    The first digit must not be zero: .9 choices.
    The second digit can be any digit: .10 choices.
    The last digit is even: .5 choices.

    Hence, there are: . (9)(10)(5) \:=\:450 numbers in set C\;\;\hdots\;\;n(C) = 450



    Now we use the inclusion-exclusion principle
    to eliminate the overlap of sets B and C.

    Their intersection:
    The 3-digit numbers that begin with {2,4,6,8} and end with {0,2,4,6,8}.
    The first digit is from {2,4,6,8}: .4 choices.
    The second digit can be any digit: .10 choices.
    The last digit is from {0,2,4,6,8}: .5 choices.

    Hence, there are: . (4)(10)(5) \:=\:200 numbers in the overlap . \hdots\;\;n(B \cap C) = 200


    Therefore: . n(B \cup C) \;=\;n(B) + n(C) - n(B \cap C) \;=\;400 + 450 -200 \;=\;\boxed{650}

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