Counting

Discussion in 'Probability and Statistics' started by Oiler, Oct 13, 2021.

  1. Oiler

    Oiler

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    Seven people enter a lift. The list stops at three floors. At each of the three floors no one enters the lift, nut at least one person leaves the lift. After the three floor stops, the lift is empty. In how many ways can this happen?
     
    Oiler, Oct 13, 2021
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  2. Oiler

    MathLover1

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    If there were no restrictions, there would be three choices for each of the seven people, so there would be 3^7 ways the seven people could depart the lift. However, at least one person gets off the lift at each floor. Therefore, we must exclude those arrangements in which fewer than three floors are used as exits.

    There are C(3,1) ways to exclude one of the floors as an exit and 2^7 ways for the people to depart the lift on the remaining floors.

    However, if we subtract C(3,1)2^7 from 3^7, we will have subtracted those distributions in which all seven people depart the lift on the same floor twice, once for each way we could have excluded one of the other floors. Since we only want to subtract these distributions once, we must add them back.

    There are C(3,2) ways to exclude two of the three floors and one way for all seven people to exit the lift on the remaining floor.

    Thus, by the Inclusion-Exclusion Principle, the number of ways seven people can depart on the lift on three floors with at least one person leaving the lift on each floor is

    3^7-C(3,1)2^7-C(3,2)1^7=1800 ways
     
    MathLover1, Oct 13, 2021
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    nycmathguy likes this.
  3. Oiler

    nycmathguy

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    Wow! You also know probability at a comfortable level. Impressive.
     
    nycmathguy, Oct 13, 2021
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