Results 1 to 5 of 5

Math Help - I'm stuck with a few combination\permutation questions. Any help?

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    Israel
    Posts
    4

    I'm stuck with a few combination\permutation questions. Any help?

    Hello.

    I'm studying for a statistics exam, and I'm stuck with a few combination\permutation questions. Here are the questions (with answers):

    Question 1:
    In a bucket there are: 4 blue balls, 5 white balls, and 6 red balls. We choose 3 balls randomly:
    a) How many choosing possibilities are there if all 3 balls are different in color? (Answer: 120)
    b) How many choosing possibilities are there if at least 2 of the 3 balls are identical in color? (Answer: 335)

    Question 2:
    In a certain language, the alphabet consists of 22 letters. How many different words consisiting of 6 characters can we form, considering each word has to contain at least 2 identical characters? (Answer: 59658544)

    Question 3:
    From a group of 7 men and 8 women, we want to choose an expedition that consists of 4 people and contains an equal number of men and women. In this expedition there are 3 jobs: leader, treasurer, and spokesman of the expedition. The fourth person of the expedition doesn't have a defined job. How many different choosing possibilites do we have in order to achieve such an expedition, if every person had the qualifications to hold any of the jobs? (Answer: 14112)

    The answers were provided with the questions, however without explanation. Can someone please explain to me in detail how to get to the answers?

    Thanks a lot in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Feb 2014
    From
    United States
    Posts
    673
    Thanks
    343

    Re: I'm stuck with a few combination\permutation questions. Any help?

    You will learn a lot more if you work these out yourself with our help. Let's try it that way. To do this kind of problem, it is essential to think about what the question means. "We choose 3 balls from a bucket containing balls of 3 colors and all three balls are of a different color." What means is that 1 ball is blue, 1 ball is white, and 1 ball is red. Given that there are 4 blue balls, how many different ways can I choose one of them? How many ways can I choose 1 white ball out of 5 possible choices? So for each single choice of a blue ball, I have how many choices for a white ball? So how many choices do I have overall for 1 blue ball and 1 white ball? Now think about the red ball. How many ways to pick one out of a collection of six? Now for each of the ways to pick a red ball, what have we already determined to be the number of ways to pick one blue ball and one white ball?

    Answer as many of those questions as you can. If you can answer all of them, you are one step away from answering 1a.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,931
    Thanks
    782

    Re: I'm stuck with a few combination\permutation questions. Any help?

    For questions like this, you should consider all possible outcomes. Here are the outcomes for the first problem:
    You have three balls, each a different color.
    You have at least two balls of the same color.
    An event and its opposite are mutually exclusive. Since there are a total of 15 balls, the total number of choosing possibilities is the number of ways to choose 3 balls from a bucket of 15 balls. Since each outcome is distinct and constitute all possible outcomes, the sum principle says that the sum of the choosing possibilities for these two will give you the total number of choosing possibilities.

    Same thing goes for question 2. Consider the opposite of what it is asking. Which is easier to count? In this case, the opposite is easier to count. How many six-character words contain no two identical characters? That is equivalent to asking how many six-character words contain six distinct characters? Subtract that from the total number of possible words (with no restrictions) to find the number of different words the problem is asking.

    For question 3, try breaking the problem down into easier pieces. For any chosen group, you can line the members up and assign jobs to the first three people in line in order: leader, treasurer, and spokesman. The fourth group member would not have a job. So, the number of ways to choose the group is independent from the number of ways to choose the jobs for the group members. This allows you to use the product principle. It is the number of ways to choose the members of the group times the number of ways to give the jobs to the members.
    Last edited by SlipEternal; July 10th 2014 at 09:16 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2012
    From
    Israel
    Posts
    4

    Re: I'm stuck with a few combination\permutation questions. Any help?

    Quote Originally Posted by SlipEternal View Post
    For questions like this, you should consider all possible outcomes. Here are the outcomes for the first problem:
    You have three balls, each a different color.
    You have at least two balls of the same color.
    An event and its opposite are mutually exclusive. Since there are a total of 15 balls, the total number of choosing possibilities is the number of ways to choose 3 balls from a bucket of 15 balls. Since each outcome is distinct and constitute all possible outcomes, the sum principle says that the sum of the choosing possibilities for these two will give you the total number of choosing possibilities.

    Same thing goes for question 2. Consider the opposite of what it is asking. Which is easier to count? In this case, the opposite is easier to count. How many six-character words contain no two identical characters? That is equivalent to asking how many six-character words contain six distinct characters? Subtract that from the total number of possible words (with no restrictions) to find the number of different words the problem is asking.

    For question 3, try breaking the problem down into easier pieces. For any chosen group, you can line the members up and assign jobs to the first three people in line in order: leader, treasurer, and spokesman. The fourth group member would not have a job. So, the number of ways to choose the group is independent from the number of ways to choose the jobs for the group members. This allows you to use the product principle. It is the number of ways to choose the members of the group times the number of ways to give the jobs to the members.
    Can you please solve the questions for me? I tried but couldn't get to the answer... Statistics just doesn't enter my brain.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,931
    Thanks
    782

    Re: I'm stuck with a few combination\permutation questions. Any help?

    Could you post your thoughts for each? You already have the answers for each. I know how to solve each, but I don't know what you are trying when you try to solve them. Try to be as descriptive as possible. Hopefully, we can figure out how to steer your brain in the right direction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutation and Combination
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: July 7th 2009, 08:44 PM
  2. Replies: 1
    Last Post: March 28th 2009, 01:12 AM
  3. Permutation or Combination?
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 16th 2008, 10:33 AM
  4. permutation and combination
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 25th 2008, 07:47 AM
  5. Permutation and Combination Help!
    Posted in the Statistics Forum
    Replies: 6
    Last Post: March 18th 2008, 12:21 PM

Search Tags


/mathhelpforum @mathhelpforum