Combinatorios and pyramids

Discussion in 'Advanced Applied Math' started by Loulou, May 2, 2022.

  1. Loulou

    Loulou

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    Hello everybody,
    I am trying to solve this problem but I need help:
    Suppose you have a pyramid (built with logs for example). You can only add a log if there are already two logs underneath and each new floor of the pyramid has one less log than the one before. The question is: for a pyramid with n logs on the first floor ( and n-1 logs on the second floor etc), on how many orders can the pyramid be built? n can be any natural numbers.
    For instance, if the base of the pyramid is 2, you can only build it in 2 ways. For the first log, you have the choice between two places on the first floor, then the second one has to be the other one on the first floor, and the third one has to be the only one on the second floor and that's it.
    Thank you very much, I can't solve it for n.
     
    Loulou, May 2, 2022
    #1
  2. Loulou

    RobertSmart

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    To solve this problem, let's denote the number of ways to build a pyramid with n logs on the first floor as P(n).

    For the base case, P(1) = 1, because there's only one way to build a pyramid with one log.

    Now, consider building a pyramid with n+1 logs on the first floor. The first log can be placed in any of the n+1 positions on the bottom layer. Once the first log is placed, there are two logs underneath it, and we need to place the remaining n logs.

    If we place the next log directly on top of the first log, then we have a subproblem of building a pyramid with n logs on the top floor. If we place the next log on one of the two logs underneath the first log, then we have a subproblem of building a pyramid with n-1 logs on the top floor.

    Therefore, the total number of ways to build a pyramid with n+1 logs on the first floor is the sum of the number of ways to build pyramids with n logs and n-1 logs on the top floor, for each possible position of the first log on the bottom layer.

    Mathematically, this can be expressed as: (+1)=()+(−1)+(−2)+…+(1)P(n+1)=P(n)+P(n−1)+P(n−2)+…+P(1)

    So, the number of orders to build a pyramid with n logs on the first floor can be calculated recursively using the above formula.

    By the way, if you need further assistance with math assignments or want to explore more problems, you might find useful resources at website of MathsAssignmentHelp. You can contact them at +1 (315) 557-6473.
     
    RobertSmart, Apr 24, 2024
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