Results 1 to 3 of 3

Math Help - Combinations and a bit of probability

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Combinations and a bit of probability

    Another set of interesting questions:

    1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?

    2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.

    3. 25 people sit around a circular table. Three of them are chosen randomly. What is the probability that at least two of the three are sitting next to one another?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by usagi_killer View Post
    1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?
    {8\choose2} = 28 ways of choosing which two rows, same number of ways of choosing which two columns, total of 28^2 = 784 rectangles.

    That assumes that the rectangles have sides parallel to the sides of the board. If you are not making that assumption then there will be many other possibilities, such as when the four squares are a knight's move from their neighbours.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.
    Here is how I would approach this one. There are always different approaches to counting problems.

    Arrange the Maples and Oaks in 7!=5040 ways. This now gives us 8 spaces in between to put the birch trees so they are separated and not next to one another. We can choose these spots in
    C(8,5)=56 ways and 5!=120 ways to arrange them.

    There is 12! ways to arrange them without restriction.

    So, the probability of no birch being next to another is

    \frac{\binom{8}{5}\cdot 5!\cdot 7!}{12!}=\frac{7}{99}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations/Probability
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 5th 2011, 12:24 PM
  2. Combinations and Probability
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: April 5th 2010, 12:28 AM
  3. Combinations in probability
    Posted in the Statistics Forum
    Replies: 3
    Last Post: February 23rd 2009, 02:25 AM
  4. Combinations and Probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 1st 2008, 07:58 PM
  5. Combinations and Probability
    Posted in the Statistics Forum
    Replies: 4
    Last Post: February 18th 2008, 12:01 PM

Search Tags


/mathhelpforum @mathhelpforum