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Homework Help: 2 little reasoning problems

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data

    2.10. Express each of the following statements as a conditional statement in "if-then" form or as a universally quantified statement. Also write the negation (without phrases like "it is false that").

    a) Every off number is a prime
    b) The sum of the angles of a triangle is 180 degrees
    c) Passing the test requires solving all the problems
    d) Being first in line guarantees getting a good seat
    e) Lockers must be turned in by the last day of class
    f) Haste makes waste
    g) I get mad whenever you do that
    h) I won't say that unless I mean it

    2.41. A clerk returns n hats to n people who have checked them, but not necessarily in the right order. For which k is it possible that exactly k people get a wrong hat? Phrase your conclusion as a biconditional statement.

    2. Relevant equations

    ?

    3. The attempt at a solution

    The only part of the first problem that I'm stuck on is b.

    The sum of the angles of a triangle is 180 degrees

    What does "a" triangle mean? Does that mean "any" triangle or "one" triangle? I don't know how to interpret that.

    And the second problem is just plain confusing.

    So, suppose n=4.

    You could have k=1,2,3 or 4.

    You could have

    right, wrong, wrong, wrong
    right, right, wrong, wrong
    right, right, right, right
    .
    .
    .

    ... Is that the right idea? I'm not sure if I'm doing this right.
     
  2. jcsd
  3. Jun 30, 2010 #2

    LCKurtz

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    It means "any".

    Could exactly 1 person have the wrong hat? Is there something special about 1?
     
  4. Jun 30, 2010 #3

    Mark44

    Staff: Mentor

    What's an "off" number? Do you mean odd?
    "A" triangle means any old triangle. All you can assume is that it has three sides.
    If n = 4, is it possible for exactly one person to get the wrong hat? It's certainly possible for two people to get the wrong hat.

    You won't be able to assume any specific value for n, but I think it's useful to look at different cases - n = 1, n = 2, n = 3, n = 4, etc., to see if you can detect a pattern.

    Also, don't forget k = 0.
     
  5. Jun 30, 2010 #4
    No. 3 persons have the right hat ----> 4 persons have the right hat. I don't think it's the same way for 3 persons having the wrong hat ...

    Anyways, why does this matter?
     
  6. Jun 30, 2010 #5

    Mark44

    Staff: Mentor

    Because the problem asks you "For which k is it possible that exactly k people get a wrong hat? "
     
  7. Jun 30, 2010 #6

    LCKurtz

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    Why it matters is you are trying to decide for what values of k, exactly that many people can have the wrong hat. And you have apparently decided it can't be k = 1. What about k = 2, or 3, or...?
     
  8. Jun 30, 2010 #7
    n=1

    You can only have one person get a wrong a hat or one person get the right hat. k=1.

    n=2

    You can have two persons get the wrong hat or you can have two person get the right hat. k=2.

    n=3

    You can have two persons get the wrong hat and one person get the right hate; or you can have everyone get the wrong hat. k=2,3.

    n=4

    You can have two persons get the right hat and the other two get the wrong hat; or you can have everyone get the wrong hate; or you can have 3 persons get the wrong hat and one get the right hat.

    k=2,3,4.

    .
    .
    .
    .
    .
    .

    Is there a pattern I'm supposed to be detecting?
     
    Last edited: Jun 30, 2010
  9. Jun 30, 2010 #8

    Mark44

    Staff: Mentor

    You're skipping a lot of cases, or maybe just not being clear about what you're saying.
    n = 1. How can one person with one hat get the "wrong" hat. There is only one hat.
    n = 2. What are the circumstances for k = 0, k = 1, k = 2?
    And so on.
     
  10. Jun 30, 2010 #9
    n=1

    You can only have one person get a wrong a hat or one person get the right hat. k=1.

    n=2

    You can have two persons get the wrong hat or you can have two person get the right hat. k=1.

    n=3

    You can have two persons get the wrong hat and one person get the right hate; or you can have everyone get the right hat. k=1,3.

    n=4

    You can have two persons get the right hat and the other two get the wrong hat; or you can have everyone get the right hat. k=2,4.

    .
    .
    .
    .
    .
    .

    Is there a pattern I'm supposed to be detecting?
     
  11. Jun 30, 2010 #10
    n=2

    Either one gets the right hat and one gets the wrong hat; or both get the right hat.

    k=2.

    n=3

    Either two get the wrong hat and one gets the right hat; or all get the wrong hat.

    k=2,3.
     
  12. Jun 30, 2010 #11

    Mark44

    Staff: Mentor

    Again, how can one person get the wrong hat if there is only one hat?

    Here's the situation. A guy walks into a restaurant where there are no other patrons, and checks his hat at the hat-check stand. After he checks it in, there is only one hat. After he finishes eating and pays for his meal, he goes back to the hat-check. How can he possibly get the wrong hat?
    What does k = 1 mean here? The question you are supposed to be answering is "For which k is it possible that exactly k people get a wrong hat?" Is it possible for 1 person to get the wrong hat where there are two people and two hats?
    You're not considering all the possibilities. For example, 3 people can get the wrong hat and 1 person the right hat. Here's one possibility:
    Person: 1 2 3 4
    Hat....: 1 3 4 2

    Person 1 gets the right hat, person 2 gets hat 3, person 3 gets hat 4 and person 4 gets hat 2.
     
  13. Jun 30, 2010 #12
    Okay ... Let me try this one last time.

    n=1

    Person: 1
    Hat: 1

    n=2

    Person: 1 2
    Hat: 1 2


    Person: 1 2
    Hat: 2 1

    So either they both get the wrong hat or they both get the right hat.

    n=3

    Person: 1 2 3
    Hat: 1 2 3

    All right

    Person: 1 2 3
    Hat: 1 3 2

    1 right, 2 wrong

    Person: 1 2 3
    Hat: 2 1 3

    1 right, two wrong

    Person: 1 2 3
    Hat: 2 3 1

    all wrong

    Person: 1 2 3
    Hat: 3 1 2

    all wrong

    Person: 1 2 3
    Hat: 3 2 1

    1 right, 2 wrong

    n=4

    Person: 1 2 3 4
    Hat: 1 2 3 4

    Person: 1 2 3 4
    Hat: 1 2 4 3

    Person: 1 2 3 4
    Hat: 1 .............

    .
    .
    .
    .
    .
    .

    etc.

    Is brute force the way to do this?

    k can be 3 or 2 when n=3 and only 2 when n=2.

    I don't understand the phrasing "For which k is it possible that exactly k".

    Can someone explain that?
     
  14. Jun 30, 2010 #13

    Mark44

    Staff: Mentor

    How else other than "brute force" would you analyze this situation? You're only using brute force, as you call it, to get an idea of what is going on for relatively small values of n.

    Now that you have looked at the various cases for n = 1, 2, 3, and 4, summarize what you have found, stating the values of k for which exactly k people get the wrong hat.

    k is an integer such that 0 <= k <= n.
    So, for n = 1, is it possible that 0 persons get the wrong hat? Is it possible for 1 person to get the wrong hat?

    For n = 2, is it possible that 0 persons get the wrong hat? Is it possible for 1 person to get the wrong hat? Is it possible for 2 persons to get the wrong hat?

    Etc.

    A table might be a way to display this information.
     
  15. Jun 30, 2010 #14
    n=1

    0/1! chance of 1 wrong hat

    n=2

    1/2! chance of 0 wrong hats
    0/1! chance of 1 wrong hat
    1/2! chance of 2 wrong hats

    n=3

    1/3! chance of 0 wrong hats
    0/3! chance of 1 wrong hat
    3/3! chance of 2 wrong hats
    2/3! chance of 3 wrong hats.

    n=4

    1/4! chance of 0 wrong hats
    0/4! chance of 1 wrong hat
    5/4! chance of 2 wrong hats
    9/4! chance of 3 wrong hats
    9/4! chance of 4 wrong hats

    .
    .
    .


    So, for n persons there is always an 1/n! chance of getting the hats all right and a 0 chance of getting just one hat wrong. Other than that, I don't see a pattern heating up.

    And I still don't understand the question.

    Maybe the chance of getting 2 hats wrong has a pattern? I see that when n=2, the chance was 1/2!; when n=3, the chance was 3/3!; when n=4, the chance was 5/4!. Maybe if j represents the chance of getting 2 wrong, j=(2n-3)/n!

    ????
     
  16. Jun 30, 2010 #15

    Mark44

    Staff: Mentor

    You're carrying this further than I would have, by counting the number of ways something can happen. All that's needed, I believe, is whether something can happen, not how many ways it can happen. This isn't a probability problem.
     
  17. Jun 30, 2010 #16
    I just don't understand how to construct my answer.

    "It is possible that exactly __ people get a wrong hat."

    ?

    And in the blank would be some expression involving n?

    Give me an example of an answer but don't use correct values.
     
  18. Jun 30, 2010 #17

    Mark44

    Staff: Mentor

    I think it's too soon to jump to the answer for this problem. It might be useful to look at a couple more cases, n = 5 and n = 6. That might give more insight to this problem, such as does it matter whether n is even or odd.
     
  19. Jun 30, 2010 #18
    5! = 120. You kidding? Show me the non-brute force way, or at least lead in the right direction.

    Whenever you think in terms of groups of 3, say 7 (3+3+1) or 12 (3+3+3+3), you have groups where either all 3 have wrong hats, 2 have wrong hats, or 0 have wrong hats in those sub-groups.

    Whenever you have groups of 2 persons, you have either 2 wrong hats or zero wrong hats.

    Do I proceed from there? I know I can think of n in therms of 2 and 3.
     
  20. Jun 30, 2010 #19

    LCKurtz

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    Here's another hint. Regardless of what n is but greater than 1, if you pick a subset of the people with k at least 2 people, can you see a way each of them could pass their hat along to the next so they all have wrong hats while all the others has their own hat?
     
  21. Jun 30, 2010 #20
    In the blank would be "k."

    Which values of k are allowed for any given value of n?
    Which values of k are not allowed?
     
  22. Jun 30, 2010 #21

    Mark44

    Staff: Mentor

    Again, you don't have to count all of the ways.
     
  23. Jun 30, 2010 #22
    The "it is possible" seems like a weak statement to me.

    What is this problem getting at? I need more hints!
     
  24. Jun 30, 2010 #23
    Look carefully at the problem, and consider what you are supposed to be asking.

    "Is it possible" is equivalent to "does at least one example exist"--why would you have to enumerate every example for a given value of k, when it is sufficient to find one?
     
  25. Jun 30, 2010 #24
    The problems was:

    A clerk returns n hats to n people who have checked them, but not necessarily in the right order. For which k is it possible that exactly k people get a wrong hat? Phrase your conclusion as a biconditional statement.

    I think:

    "There exists a number k≤n representing the number of people who get a wrong hat"

    Okay, well I know k≠1 because if (n - 1) persons get the right hat, then n persons get the right hat.

    That's all I've established.
     
  26. Jun 30, 2010 #25
    There is not a single number k for most values of n. This is one point where you are going wrong.

    Plus, you don't need to consider all n! permutations of n people with hats.
    Partition the group into people with the incorrect hat, and people wearing the correct hat.

    Now consider this: If the hat I have is not my own, is its owner wearing the correct hat?
     
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