Results 1 to 4 of 4

Thread: Permutation trouble

  1. #1
    Newbie
    Joined
    Jun 2018
    From
    Netherlands
    Posts
    2

    Permutation trouble

    Hello,

    I am wondering whether I solved the following correctly.

    "How many license plates containing seven characters can be created from letters and numbers if the first two characters must be uppercase letters and the last five characters must be digits from 0 to 9? Repetition of numbers is allowed. Repetition of letters is not allowed".

    Here's my solution.

    1) I assumed a universal set for the license plate as U = {uppercase letter, uppercase letter, number, number, number, number, number}.
    2) For the first two characters I assumed a sample set A ={A,B,C...Z}, for the last five characters there's the sample set B = {0,1,2,3...9}.
    3) Considering the repetition of letters is not allowed and that the order matters in set U, I determined 650 possible permutations in set A, following the formula n!/(n-r)!, which comes down to 26!/(26-2)! = 650 permutations. So set A has 650 possible permutations.
    4) Considering the repetitions of numbers is allowed and that the order in set U matters, I determined 100.000 possible permutations in set B, following the formula n^r, which comes down to 10^5 = 100.000 permutations. So set B has 100.000 possible permutations.
    5) Now set A and B are added together, giving what I think are 100.650 possible license plates.

    Are my thoughts correct or did I go wrong somewhere? It feels like I went wrong somewhere. Thanks in advance =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,042
    Thanks
    2957
    Awards
    1

    Re: Permutation trouble

    Quote Originally Posted by Wolfy1984 View Post
    "How many license plates containing seven characters can be created from letters and numbers if the first two characters must be uppercase letters and the last five characters must be digits from 0 to 9? Repetition of numbers is allowed. Repetition of letters is not allowed".

    Here's my solution.

    1) I assumed a universal set for the license plate as U = {uppercase letter, uppercase letter, number, number, number, number, number}.
    2) For the first two characters I assumed a sample set A ={A,B,C...Z}, for the last five characters there's the sample set B = {0,1,2,3...9}.
    3) Considering the repetition of letters is not allowed and that the order matters in set U, I determined 650 possible permutations in set A, following the formula n!/(n-r)!, which comes down to 26!/(26-2)! = 650 permutations. So set A has 650 possible permutations.
    4) Considering the repetitions of numbers is allowed and that the order in set U matters, I determined 100.000 possible permutations in set B, following the formula n^r, which comes down to 10^5 = 100.000 permutations. So set B has 100.000 possible permutations.
    5) Now set A and B are added together, giving what I think are 100.650 possible license plates.
    The results are multiplied not added: $(650)\cdot(10^5)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,280
    Thanks
    2686

    Re: Permutation trouble

    For the first letter you have 26 possible choices. For the 2nd letter you have 25.

    For each of the numbers you have 10 choices. There are 5 of these.

    so the total number of unique license plates is given by

    $N = 26 \cdot 25 \cdot 10^5 = 65,000,000$

    Ok... I just read through your work.... you did so well until you decided to add instead of multiply. Think about it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2018
    From
    Netherlands
    Posts
    2

    Re: Permutation trouble

    Ahh I get it, set U is a simple product of sets A and B. Thanks for the replies =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. What is the permutation of pi?
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Dec 15th 2012, 05:36 PM
  2. Is this a permutation?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Apr 4th 2011, 12:43 AM
  3. permutation 2
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 8th 2010, 07:01 AM
  4. permutation
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Aug 9th 2009, 08:54 AM
  5. not a permutation? then how?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Mar 16th 2008, 12:50 AM

/mathhelpforum @mathhelpforum