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Math Help - counting and permutation

  1. #1
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    counting and permutation

    Given the digits 1,2,3,4 and5 find how many 4-digit numbers can be formed from them:

    (a)if the number must be even,without any repeated digit
    (b)if the number must be even
    (c)if repetitions of a digit are allowed

    i'm not sure in my answer. Please answer to me.
    thank you.
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  2. #2
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    Please show us what you think the answers should be.
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by calfever View Post
    Given the digits 1,2,3,4 and5 find how many 4-digit numbers can be formed from them:

    (a)if the number must be even,without any repeated digit
    (b)if the number must be even
    (c)if repetitions of a digit are allowed

    i'm not sure in my answer. Please answer to me.
    thank you.
    I'll start you off, for a) if the number is even it must end in 2 or 4.

    Then you would have two cases of a permutation of 3 digits taken from 4 numbers. Like Plato said, attempt the rest, show your work, and you will get excellent help.
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  4. #4
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    (a)4*3*2*2
    (b)5*5*5*2
    (c)5*5*4*3

    correct?
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  5. #5
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    1.how many natural numbers greater than or equal to 1000 and less than 5400 have the properties:
    (a)no digit is repeated
    [my ans.4(9*8*7)+(4*5*4)]
    (b)the digits 2 and 7 do not occur
    [my ans.3(7*7*7)+1+(3*4*4)

    2.how many 6-digit numbers can be formed using {1,2,...,9}with no repetitions such that 1 and 2 do not occor in consecutive positions?
    [my ans.(9*8*7*6*5*4)-(7*6*5*4*5)

    3.how many positive integers less than 1,000,000 can be written using only the digits 7,8 and9? how many using only the digits 0,8 and9?
    [my ans.3^1+3^2+...+3^6] and [my ans.3+2(3+3^2+...+3^5)]

    Please check my answer. thank you.
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  6. #6
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by calfever View Post
    (a)4*3*2*2
    (b)5*5*5*2
    (c)5*5*4*3

    correct?
    Not even close.

    Remember, I told you that for a) you had two cases:

    First case the number ends in 2 so you have \frac{4!}{(4-3)!}
    Second case the number ends in 4 so you have \frac{4!}{(4-3)!}

    When you have two mutually exclusive cases for one event you add the two together giving 4! + 4!

    Part b) is totally different. The numbers can be repeated. So for the first digit you have 5 choices, the second digit you have 5 choices, the third digit you have five choices, but for the fourth digit you only have 2 choices since the number must be even. How would you calculate that?
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