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Repeat the Path for 10 Doors and 8 Staircases

  1. Aug 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Josh works on the second floor of a building. There are 10 doors to the building and 8 staircases from 1st to the 2nd floor. Josh decided that each day he would enter by one door and leave by a different one and go up one stair case and down another. How many days could Josh do this before he had to repeat a path he had previously taken?

    2. Relevant equations


    3. The attempt at a solution
    For one day: Possibilities for choosing a door are 10 * 9
    For one day: Possibilities for choosing a Stair case are 8 * 7

    I dont know after how many days he would repeat the path.

    Zulfi.
     
  2. jcsd
  3. Aug 9, 2017 #2

    berkeman

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    Is the limiting factor the number of doors or the number of staircases?
     
  4. Aug 9, 2017 #3
    Hi,
    Both are limited. Doors are 10 and staircases are 8. Or you can say that there are less stair cases so he would repeat the stair case path early.

    Zulfi.
     
  5. Aug 9, 2017 #4

    berkeman

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    Exactly. And with 8 staircases that you are choosing up and down from, how many days can you go without repeating the same up/down choice? Describe how you would figure that out... (how do count the combinations...)

    EDIT -- Please see later posts...
     
    Last edited: Aug 11, 2017
  6. Aug 9, 2017 #5

    SammyS

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    Ask yourself: "What does a path consist of ?"
     
  7. Aug 10, 2017 #6

    LCKurtz

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    I don't think this question involves anything more than the so called "sequential counting principle". Upon entering the building you have 10 choices for a door. After that you have __ choices for a stairway to enter. After that you have __ choices for a stairway to leave. After that you have __ choices for an exit door. So...
     
  8. Aug 10, 2017 #7

    berkeman

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    Ah, interesting. I just realized that I misread the question. If you include the doors as variations on the overall path choices, that does increase the final number. Thanks.
     
  9. Aug 11, 2017 #8
    Hi,

    There are more doors and less stair cases. So we would stop when we would have a repeatation of Staircases. Answer should be in factorial. 10! for Doors which is the number of days but before these number of days we would stop because there are less number of stair case . So we have 10!/8!. Or it would 8!/2! because we have 2 staircases per day.

    Somebody please guide me.

    Zulfi.
     
  10. Aug 11, 2017 #9

    SammyS

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    Neither of those.

    Consider what @LCKurtz wrote.

    There are many, many different "paths" which are possible.
     
  11. Aug 12, 2017 #10
    Hi,

    I think LCKurtz talks about Counting Principle as in the book. Yes its possible: For Doors he has : 10 * 9 options or days i.e after 90 days there is a possibility that he is going to perform a repeat enterance from the same and leave from the same i.e. the duo is same as he did once in the last 90 days. Let he has has Doors D1..D10.

    So first day he entered through D1 & leave through D2. Second day he entered again through D1 but left through D3. Thus for 1st nine days he kept entering from D1 but choose other 9 doors for leaving each time he left with a unique door. On 10th day he entered through D2 but choose D1 & D3..D10 for leaving, keeping in view the uniqueness factor for the next 9 days. Thus on 91st day there is a possibility of repeatation. That is he enters from D1 & leaves through D2. So we wont consider the remaining options.


    For Stair cases, in the same way, on the first he can chose from any of the 1-8 staircases(S1-S8) to climb up & go down through1-7 staircases in order to avoid repeatation. Let suppose he went up through S1 & came down through S2. On the second day he went up through S1 & came down through S3. So he has following combinations with out repeatation for first 7 days: {S1-S2, S1-S3, S1-S4, S1-S5,S1-S6, S1-S7, S1-S8} On the 8th day he can choose S2 fopr up & S3 for down so he has following climbing pattern for next seven days:{S2-S1, S2-S3, S2-S4, S2-S5, S2-S6,S2-S7, S2-S8}. Thus for first 56 days no repeation occurs but 57 day, there is a repeatation.

    If we limit by staircases the he can have only 56 days without repeatation but if we consider the whole path from Door In- Stair Up-Stair Down- Door out then he can have:

    10 * 9 * 8 * 7 days with repeatation of any path.


    This is the answer in the book also.


    Thanks everybody.


    Zulfi.
     
  12. Aug 12, 2017 #11

    LCKurtz

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    Correct. But you don't need all that verbiage. Just fill in the blanks in post #6 and write the answer down.
     
  13. Aug 12, 2017 #12
    Hi,
    Actually it would be useful for my future understanding. I have stored the link.

    Zulfi.
     
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