Results 1 to 6 of 6

Math Help - Permutations question

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    canada
    Posts
    45

    Permutations question

    1. How many numbers of at most three digits can be created from the digits 0,1,2,3, and 4? I keep getting 68, answer is apparently 69.

    i did , 4*4*3 + 4*4 + 4 = 68

    2.How many 4 digit even numbers greater than 5000 can you form using the digits 0,1,2,3,5,6,8 and 9 without repetitions? How do i go about this?

    b) how many of these numbers end in 0?

    im doing the cases, and i set it up as 1*7*6*2 + 3*7*6*3 which isnt right


    n_P_3=120 n!/(n-3)! = 120 = n(n-1)(n-2)=120 doesnt make sense from here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,812
    Thanks
    661

    Re: Permutations question

    Hey skg94.

    For 1) can we have any number that is 1, 2, or 3 digits given the five possibilities? If this is the case, then I think you are underestimating the possibilities.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    75
    Thanks
    1

    Re: Permutations question

    I approached 1) by considering the quantity of single digit numbers that can be made using the five possibilities. That's just n_P_r where n=5 and r=1. I thought 5_P_r might represent the quantity of r-digit numbers that can be made using five possibilities (which it does) so I figured if I added the total quantity of numbers that can be made from r=1 to r=3, then I'd get the solution to your problem. That is:

    5_P_1 + 5_P_2 + 5_P_3 which gave me 85.

    I then noticed that for r=2: 01=1, 02=2, 03=3, 04=4 and for r=3: 012=12, 013=13, 014=14, 021=21, 023=23, 024=24, 031=31, 032=32, 034=34, 041=41, 042=42, 043=43. Since they represent the same number, these 16 possibilities must be discarded.

    85-16=69



    I'm not sure if this is the only way to arrive at the solution since I did out by hand all the possibilities which I thought would repeat (even though I knew each of n_P_r, n_P_(r+1) etc... for r>1 had distinct repeats). Perhaps if I figured out a more general rule that tells me how many repeats I have for each "r" when r>1, then I wouldn't have to list the possibilities for repeating numbers and in doing so, I would arrive with the complete solution.

    Am I on the right track here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133

    Re: Permutations question

    Hi Skg94

    For (1), is it that you are not counting the number zero as a number, (your third component should be 5). ?

    For (2), the possible forms for the four digits are,
    (Odd, 5 or 9) (any one of 6) (any one of 5) (even, 0, 2, 6 or 8).................8 times 30 possibles.
    (Even, 6 or 8) (any one of 6) (any one of 5) (even, 0, 2, one of (6 or 8))............6 times 30 possibles.
    Of these, it looks like there are four basic forms ending with zero, 5**0, 9**0, 6**0 and 8**0, with 30 possibles for each.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    694

    Re: Permutations question

    Hello, skg94!

    2. How many 4-digit even numbers greater than 5000
    can you form using the digits 0,1,2,3,5,6,8 and 9 without repetitions?

    There are 4 even digits: {0, 2, 6, 8} and 4 odd digits: {1, 3, 5, 9}.

    The first digit can be {5, 6, 8, 9}.

    If the first digit is odd: 5 or 9 . . . 2 choices
    . . the last digit can be: 0, 2, 6, 8 . . . 4 choices
    The other two digits have (6)(5) = 30 choices.
    Hence, there are: (2)(4)(30) = 240 such numbers.

    If the first digit is even: 2, 6, or 8 . . . 3 choices
    . . the last digit has 3 choices . . . 3 choices
    The other two digits have (6)(5) = 30 choices.
    Hence, there are: (3)(3)(30) = 270 such numbers.

    Therefore, there are: 240 + 270 \,=\,\boxed{510} even numbers greater than 5000.




    b) How many of these numbers end in 0?

    The last digit is 0 (zero) . . . 1 choice.
    The first digit is 5, 6, 8, or 9 . . . 4 choices
    The other two digits have (6)(5) = 30 choices.

    Therefore, there are: (1)(4)(30) = \boxed{120} such numbers.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    659
    Thanks
    271

    Re: Permutations question

    Hi Skg94,
    Counting problems are hard -- don't let anyone tell you otherwise. As illustration, the answer to 2 given by Soroban is wrong (he did come up with the right number for the second part).
    Any 4 digit number is of the form 1000d + 100c + 10b + a. Choose the digits in the order d, a, b, c.
    For d=5, there are 4*6*5 = 120 choices
    For d=7, there are 4*6*5 = 120 choices
    For d=6, there are 3*6*5 = 90 choices (when d=6, the only choices for a are 0, 4 and 8)
    For d=8, there are 90 choices.
    So a grand total of 240+180 = 420.
    Exactly similarly, you can count the number with 0 as the least significant digit and get 120.

    If you're a computer programming whiz, you can check answers to such "small" problems with a little computer program. I did so for this problem and generated all the 4 digit integers in the question. If you know programming, I'd be happy to send you the program as a private message. See the attachment.
    allints.txt
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutations Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 13th 2010, 03:24 PM
  2. Permutations Question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 21st 2009, 04:30 AM
  3. permutations question #2
    Posted in the Statistics Forum
    Replies: 5
    Last Post: May 15th 2009, 10:39 AM
  4. Permutations question...
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 26th 2009, 02:13 PM
  5. Permutations Question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: June 30th 2008, 07:59 PM

Search Tags


/mathhelpforum @mathhelpforum