Results 1 to 7 of 7

Math Help - Question on Combinatorics

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    1

    Question on Combinatorics

    I'm learning for my highschool finals (in the Netherlands) and reviewing all the chapter I've done this and last year in maths.
    For three weeks I'll be repeating Combinatorics and Probability Theory. And I have a question my math teacher couldn't answer.

    Let's assume we need to make a row of 5 pencils. And those pencils can be black or white.
    The ammount of colour combinations I can make is 2^5

    Now suppose we can choose out of three kinds of colours: white, black and green.
    The ammount of colour combinations is 3^5

    Now let's say there must be three black pencils and two white ones
    The ammount of colour combinations is:

    5 ncr 3 or 5 ncr 2
    ( with ncr this is what I mean: 5 ncr 3 - Wolfram|Alpha)


    Now my question:
    Let's assume we have 3 black penciles, 5 white peciles and 2 green one. How many combinations can I make?

    I know you can write all possibilities down but I'm curious if there is a mathematical way and if not why not.

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1

    Re: Question on Combinatorics

    Quote Originally Posted by MuddyMudskipper View Post
    Now my question:
    Let's assume we have 3 black penciles, 5 white peciles and 2 green one. How many combinations can I make?
    I know you can write all possibilities down but I'm curious if there is a mathematical way and if not why not.
    Think of alpha-string BBBWWWWWGG there are \frac{10!}{3!\cdot5!\cdot 2!} ways to rearrange that string.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2011
    Posts
    58

    Re: Question on Combinatorics

    @Plato - I believe the question is concerned with only strings of length 5. Here is my (unsatisfactory) solution to the problem.

    We do not need to write out each possible unique 5 character string, we only need 1 example of each and then we can use counting methods. So to get an example of each, consider the fact that for each string we can either have 0,1,2, or 3 b's, then given how many b's we can either have 0,1, or 2 greens.

    So here are the examples of each string:
    WWWWW
    WWWWG
    WWWGG
    BWWWW
    BWWWG
    BWWGG
    BBWWW
    BBWWG
    BBWGG
    BBBWW
    BBBWG
    BBBGG

    Now we count the unique permutations of each of these and sum to get \frac{5!}{5!}+\frac{5!}{4!}+\frac{5!}{3!\cdot 2!}+\frac{5!}{4!}+\frac{5!}{3!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{2!\cdot 3!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{2!\cdot 2!}+\frac{5!}{3!\cdot 2!}+\frac{5!}{3!}+\frac{5!}{3!\cdot 2!}

    Like I said, I am definitely not happy and would love to see a more elegant solution, but this sum is equal to 181
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2011
    Posts
    58

    Re: Question on Combinatorics

    I know I at least think that I've "solved" this question, but can anyone see a more elegant way of counting this up? I feel like mine is stupid.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1

    Re: Question on Combinatorics

    Quote Originally Posted by MonroeYoder View Post
    I know I at least think that I've "solved" this question, but can anyone see a more elegant way of counting this up? I feel like mine is stupid.
    I am not sure that there is a simpler way than yours to solve that question if it indeed is way you read it.
    I am not convinced that is a correct reading.
    Here is my reason for thinking that: "that only complicates the solution"
    It teaches nothing new. Whereas, my reading brings a fundamentalism principle into play.
    Now yours may well be the correct reading of the question.
    But if it if then in my view it nothing more that a busy work problem.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2011
    Posts
    58

    Re: Question on Combinatorics

    I completely agree that this type of problem is "busy work." As someone who has taken Combinatorics in college I am somewhat distressed that I can't come up with a better way. I mean if we have
    x B's, y W's and z G's,
    and we want to know how many unique strings of length n we can make with the characters B W and G, it seems like there should be a general technique to do so.

    Thanks for giving it consideration though.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2012
    From
    near Seattle
    Posts
    7

    Re: Question on Combinatorics

    I am not an expert here, but why is this not a simple problem of combinations? We have 10 items taken 5 at a time
    C(10,5) = 252?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinatorics Question
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: May 2nd 2011, 03:30 AM
  2. combinatorics question
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: April 10th 2011, 10:37 AM
  3. Combinatorics question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: December 16th 2008, 10:12 AM
  4. Combinatorics Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 10th 2008, 10:21 AM
  5. Combinatorics question
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: December 8th 2007, 02:12 PM

Search Tags


/mathhelpforum @mathhelpforum