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Each person out of 6 has 6 options, hence total # of outcomes is 6^6;

Favorable outcomes will be 6!, which is # of ways to assign 6 different buttons to 6 people: 1-2-3-4-5-6 (floors) A-B-C-D-E-F (persons) B-A-C-D-E-F (persons) B-C-A-D-E-F (persons) ... So basically # of arrangements of 6 distinct objects: 6!.

Re: Six Six people are on an elevator that stops at exactly 6 [#permalink]
19 Feb 2012, 05:55

Hi, could you please tell me the mistake in the following logic?

First person entering pushes one button, it doesn't matter which: p=1 Second person pushes a button that has not been pressed before. Since one is already pushed, only 5 remain: p=5/6 Same logic for third person: p=4/6 . . .

This leaves us with The probability of all pushing a different button of: 1*5/6*4/6*4/6*2/6*1/6 or 5!/6^5

Re: Six Six people are on an elevator that stops at exactly 6 [#permalink]
19 Feb 2012, 10:48

Expert's post

Gwydion wrote:

Hi, could you please tell me the mistake in the following logic?

First person entering pushes one button, it doesn't matter which: p=1 Second person pushes a button that has not been pressed before. Since one is already pushed, only 5 remain: p=5/6 Same logic for third person: p=4/6 . . .

This leaves us with The probability of all pushing a different button of: 1*5/6*4/6*4/6*2/6*1/6 or 5!/6^5

Where's the mistake?

There is no mistake: 6!/6^6=(5!*6)/(6^5*6)=5!/6^5, the same answers. _________________

Re: Six Six people are on an elevator that stops at exactly 6 [#permalink]
07 Jun 2013, 06:52

4

This post received KUDOS

1

This post was BOOKMARKED

Gwydion wrote:

Hi, could you please tell me the mistake in the following logic?

First person entering pushes one button, it doesn't matter which: p=1 Second person pushes a button that has not been pressed before. Since one is already pushed, only 5 remain: p=5/6 Same logic for third person: p=4/6 . . .

This leaves us with The probability of all pushing a different button of: 1*5/6*4/6*4/6*2/6*1/6 or 5!/6^5

Where's the mistake?

Expanding on Gwydion's post - it is true that p=1, but because the answers all have 6 in them, the first person should be written as p=6/6 to make it easier to see the answer:

First person walks in and can push any button (6/6) Probability that second person will press any of the remaining 5 buttons (5/6) Probability that third person will press any of the remaining 4 buttons (4/6) Probability that fourth person will press any of the remaining 3 buttons (3/6) Probability that fifth person will press either of the remaining 2 buttons (2/6) Probability that sixth person will press the remaining button (1/6)

Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
10 Jul 2014, 01:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
21 May 2015, 05:15

Does this logic work ?

Assuming we are calculating the options for that one person we use the number of floors are the options. Person A presses at the ground floor has 6 options, then 5 options as so on as he goes up. so 6*5*4*3*2*1 = 6!

Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
21 May 2015, 05:51

1

This post received KUDOS

Expert's post

shallow9323 wrote:

Does this logic work ?

Assuming we are calculating the options for that one person we use the number of floors are the options. Person A presses at the ground floor has 6 options, then 5 options as so on as he goes up. so 6*5*4*3*2*1 = 6!

Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
21 May 2015, 06:10

1

This post received KUDOS

EgmatQuantExpert wrote:

shallow9323 wrote:

Does this logic work ?

Assuming we are calculating the options for that one person we use the number of floors are the options. Person A presses at the ground floor has 6 options, then 5 options as so on as he goes up. so 6*5*4*3*2*1 = 6!

Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
21 May 2015, 06:20

Expert's post

shallow9323 wrote:

EgmatQuantExpert wrote:

shallow9323 wrote:

Does this logic work ?

Assuming we are calculating the options for that one person we use the number of floors are the options. Person A presses at the ground floor has 6 options, then 5 options as so on as he goes up. so 6*5*4*3*2*1 = 6!

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