Results 1 to 4 of 4

Math Help - Combinatorial Prob.

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    12

    Combinatorial Prob.

    2 questions.

    1. Joe, an avid and properly licensed sportsman, is in his hunting blind when he locates 20 canada geese, 25 mallard ducks, 40 bald eagles, 10 whopping cranes, and 5 flamingos. Joe randomly selects six birds to target. What is the probability that at least one of each species is targeted?


    2. Thirty items are arranged in a 6 by 5 array as shown.


    Combinatorial Prob.-table.png
    Calculate the number of ways to form a set of three distinct items such that no two of the selected items are i the same row or same column.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645

    Re: Combinatorial Prob.

    Hello, kangta27!

    \text{2. Thirty numbers are arranged in a }6 \times 5\text{ array as shown.}

    . . \begin{array}{|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 \\ \hline 6 & 7 & 8 & 9 & 10 \\ \hline 11 & 12 & 13 & 14 & 15 \\ \hline 16 & 17 & 18 & 19 & 20 \\ \hline 21 & 22 & 23 & 24 & 25 \\ \hline 26 & 27 & 28 & 29 & 30 \\ \hline \end{array}

    \text{Calculate the number of ways to form a set of three distinct numbers so that}
    \text{no two of the selected numbers are in the same row or same column.}

    Choose any of the 30 available numbers.
    Cross out that number and any number in its row and in its column.
    This leaves a 5\times4 array of numbers.

    Choose any of the 20 available numbers.
    Cross out that number and any number in its row and in its column.
    This leaves a 4\times3 array of numbers.

    Choose any of the 12 available numbers.

    It seems that there are: . 30\cdot20\cdot12 \:=\:7200 ways.

    But the order of the selection of the three numbers is not considered.
    Hence, we must divide by 3! = 6

    Answer: . \dfrac{7200}{6} \,=\,1200 ways.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2011
    Posts
    83
    Thanks
    7

    Re: Combinatorial Prob.

    First question: P(\text{"At least one of each kind"})=\frac{\binom {20} 1 \binom {25} 1 \binom {40} 1 \binom {10} 1 \binom 5 1 \binom {95} 1}{\binom {100} 6}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2012
    From
    Cairo
    Posts
    2

    Re: Combinatorial Prob.

    So please how did we know that the order is not considered ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinatorial Prob.
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 11th 2012, 01:15 AM
  2. Replies: 2
    Last Post: April 12th 2011, 10:13 AM
  3. Combinatorial analysis
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 15th 2010, 05:02 AM
  4. combinatorial problem??
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: September 23rd 2009, 10:29 PM
  5. combinatorial sum
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 12th 2009, 10:46 AM

Search Tags


/mathhelpforum @mathhelpforum