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Discrete Mathematics : Counting and Probability

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Question 1:
    a) Suppose you have brought four pens of different colours to the exam. For each of the ten question on the exam, you choose one pen. In how many ways can this be done?

    b) In how many ways can you distribute six bananas and five oranges between three children so that each child receives at least one banana?

    c) Suppose you have eight apples which you wish to put in five boxes. How many possible outcomes are there if the apples are not distinguished from each other, but the boxes are?


    2. Relevant equations

    a) For this, I take you just go 10x4 = 40.

    b) I assumed you consider there to be only 10 objects after taking into account each has to have a banana, hence 10! / (10-3)! = 720

    c) I have no clue how to do this bit.


    3. The attempt at a solution

    Kindly have a look at what I have attempted and correct me if I'm wrong. Thanks!
     
  2. jcsd
  3. Nov 2, 2011 #2

    HallsofIvy

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    No. You have four choices for the first question, four choices for the second, etc. What does the "fundamental principle of counting" tell you about that?

    Ten objects? There are six bananas and five oranges, a total of 11 objects. After you assign one banana to each child there area three bananas and five oranges, a total of 8 objects.

    Since the apples are "indistinguishable", it is a question of "partitioning" the number 8. How many different ways can you find 5 non-negative integers that add to 8. The order is relevant.


    3. The attempt at a solution

    Kindly have a look at what I have attempted and correct me if I'm wrong. Thanks![/QUOTE]
     
  4. Nov 2, 2011 #3
    I'm too dense for "fundamental principle of counting" please explain. If there are 10 questions and I have 4 choices for each question, I take it you have 40 choices in total.

    For the 2nd part, it should be 8!/(8-3)! = 336

    I'm still lost with the third part prof. Thanks for helping me !!
     
  5. Nov 2, 2011 #4

    Ray Vickson

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    Look at the simpler case of 2 questions with 4 choices each. For each choice of answer on Question 1 you have 4 possible choices of answer for Question 2, so you have 4x4 = 16 choices altogether.

    RGV
     
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