My Math Forum A conceptual problem in Permutation

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 August 26th, 2013, 05:46 AM #1 Member   Joined: Oct 2011 Posts: 81 Thanks: 0 A conceptual problem in Permutation Ques) 5 Persons enter a lift cabin of a 8 storey building on ground floor. In how many ways can they all leave the lift if they can all leave at any floor(except the ground floor of course)? My Attempt- 1.)As there are seven floors at which each of the person has a choice to leave the lift on, total permutations- 7*7*7*7*7= 7^5 But then i thought that problem could also be pictured as in attempt 2. 2.) As a floor approaches, the possibility may be that i.)no one leaves ii.) one of them leave iii.) two of them leave iv.) three of them leave v.) four of them leave vi.)five(all) of them leave, which makes a case of 6 possibilities at each floor, so total permutations:- 6^7 Which is the correct method? Please guide me.
 August 26th, 2013, 05:50 AM #2 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: A conceptual problem in Permutation You're approach 2 is not complete. Assume 2 persons get out at the first floor. then there are not 6 possiblities left at fthe next floor right ? so it's not 6^7 but I don't know the right answer though.
August 26th, 2013, 05:53 AM   #3
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Re: A conceptual problem in Permutation

Quote:
 Originally Posted by gelatine1 You're approach 2 is not complete. Assume 2 persons get out at the first floor. then there are not 6 possiblities left at fthe next floor right ? so it's not 6^7 but I don't know the right answer though.
I do not know the answer myself. But yes, i guess my second approach was wrong.

 August 26th, 2013, 06:57 AM #4 Member   Joined: Apr 2013 Posts: 35 Thanks: 0 Re: A conceptual problem in Permutation I have an idea of how to solve it but first a simple question Is the probability of all of them leaving on the first floor 1/6 because there are 6 possible events and all of them leaving is one of the events or is the probability 1/16807 because every person is going to leave on one of the seven floors so the probability of person a deciding to leave on 1st floor 1/7 and since all 5 have to make that same descision 1/7^5. With that said is 16807 the number of possible optons

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