Results 1 to 8 of 8

Math Help - combination lock (question)

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    184

    combination lock (question)

    The dail on a 3 number combination lock contains markings to represent the numbers 0-59. How many combination's are possible in each case?

    a. the first and second numbers must be different, and the second and third numbers must be different.

    b. the first and second numbers must differ by atleast 3.

    I dont know how to do these.

    is it.... 3^{60}=4.24*10^{28} divided by something.........????
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by brentwoodbc View Post
    The dail on a 3 number combination lock contains markings to represent the numbers 0-59. How many combination's are possible in each case?

    a. the first and second numbers must be different, and the second and third numbers must be different.

    b. the first and second numbers must differ by atleast 3.

    I dont know how to do these.

    is it.... 3^{60}=4.24*10^{28} divided by something.........????
    If there were no restrictions on the 2nd and 3rd number then you could have 60 choices for the 1st, 2nd and 3rd.

     60 \times 60 \times 60 = 216000 which is  60^3

    or restated

     3^{11} < 60^3 < 3^{12}

    That is no where in the range of  3^{60}

    If the first number were 3, then the 2nd number could not be 1,2,3,4, or 5: you have 55 choices for the 2nd number on your lock.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2008
    Posts
    184
    thank you so
    a= 60*59*59=208860
    and
    b=60*55*60=198000? book says 198360? I get it though thanks.
    its a little tricky with the differ by three thing.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by brentwoodbc View Post
    b=60*55*60=198000? book says 198360? I get it though thanks.
    If the first setting is 0 then second cannot be 0,1,2.
    Therefore, there are 57 choices for the second setting and 60 for the third.
    The same can be said if the first is 59.

    If the first setting is 1 then second cannot be 0,1,2,3.
    Therefore, there are 56 choices for the second setting and 60 for the third.
    The same can be said if the first is 58.

    For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
    That means we can choose 55 values for the second and 60 for the third.

    (2)(57+56)(60)+(55)(56)(60)=198360. The textbook is correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2008
    Posts
    184
    Quote Originally Posted by Plato View Post
    If the first setting is 0 then second cannot be 0,1,2.
    Therefore, there are 57 choices for the second setting and 60 for the third.
    The same can be said if the first is 59.

    If the first setting is 1 then second cannot be 0,1,2,3.
    Therefore, there are 56 choices for the second setting and 60 for the third.
    The same can be said if the first is 58.

    For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
    That means we can choose 55 values for the second and 60 for the third.

    <b>(2)(57+56)(60)</b>+(55)(56)(60)=198360. The textbook is correct.
    how is it 2*(57+56)*60 I dot understand that part totally.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2008
    Posts
    184
    you also said if its 1 it cant be 0,1,2,3 , but what about if its 0?

    wouldnt it be 2*(57+57)*60 ? I dont get this.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by Plato View Post
    If the first setting is 0 then second cannot be 0,1,2.
    Therefore, there are 57 choices for the second setting and 60 for the third.
    The same can be said if the first is 59.

    If the first setting is 1 then second cannot be 0,1,2,3.
    Therefore, there are 56 choices for the second setting and 60 for the third.
    The same can be said if the first is 58.

    For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
    That means we can choose 55 values for the second and 60 for the third.

    (2)(57+56)(60)+(55)(56)(60)=198360. The textbook is correct.
    Quote Originally Posted by brentwoodbc View Post
    you also said if its 1 it cant be 0,1,2,3 , but what about if its 0?
    wouldnt it be 2*(57+57)*60 ? I dont get this.
    Do you actually read the replies?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2008
    Posts
    184
    Quote Originally Posted by Plato View Post
    If the first setting is 0 then second cannot be 0,1,2.
    Therefore, there are 57 choices for the second setting and 60 for the third.
    The same can be said if the first is 59.

    If the first setting is 1 then second cannot be 0,1,2,3.
    Therefore, there are 56 choices for the second setting and 60 for the third.
    The same can be said if the first is 58.

    For any number other than 0,1,58,59 all 56 of them, we exclude 5 possiblies of the second setting.
    That means we can choose 55 values for the second and 60 for the third.

    (2)(57+56)(60)+(55)(56)(60)=198360. The textbook is correct.
    (2)(57+56)(60)

    I just need clarification on this part
    2= number of digits where you have 57/56 choices?
    (57+56) = why do you add?

    is it 2 choices (times) where you have 57 and 56 choices multiplied by 60?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Total number of combinations for combination lock
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: December 26th 2011, 10:11 AM
  2. 9-Digit Combination Lock Chances And Probability
    Posted in the Statistics Forum
    Replies: 9
    Last Post: June 25th 2011, 03:39 PM
  3. combination lock
    Posted in the Discrete Math Forum
    Replies: 10
    Last Post: April 19th 2010, 07:26 AM
  4. Master Lock Combination
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 7th 2008, 06:22 PM
  5. Combination Lock
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2007, 03:11 PM

Search Tags


/mathhelpforum @mathhelpforum