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# Six people are on an elevator that stops at exactly 6 floors

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Six people are on an elevator that stops at exactly 6 floors [#permalink]  19 Feb 2012, 05:02
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Six people are on an elevator that stops at exactly 6 floors. What is the probability that exactly one person will push the button for each floor?

(A) 6!/6^6
(B) 6^6/6!
(C) 6/6!
(D) 6/6^6
(E) 1/6^6
[Reveal] Spoiler: OA
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Re: Six people [#permalink]  19 Feb 2012, 05:11
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caliber23 wrote:
Six people are on an elevator that stops at exactly 6 floors. What is the probability that exactly one person will push the button for each floor?

(A) 6!/6^6
(B) 6^6/6!
(C) 6/6!
(D) 6/6^6
(E) 1/6^6

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!.

P=favorable/total=6!/6^6

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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

Where's the mistake?
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Re: Six Six people are on an elevator that stops at exactly 6 [#permalink]  19 Feb 2012, 05:57
Totol possibilities: 6^6 (Each person may push each level the button)

First floor: 6 person
second floor: 5 person
...
= 6!

Possibility: 6!/6^6 --> A
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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.
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]  07 Jun 2013, 06:03
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

$$\frac{6}{6} * \frac{5}{6} * \frac{4}{6} * \frac{3}{6} * \frac{2}{6} * \frac{1}{6} = \frac{6*5*4*3*2*1}{6*6*6*6*6*6} = \frac{6!}{6^6}$$

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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]  10 Jul 2014, 01:31
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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!
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]  21 May 2015, 05:51
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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!

Dear shallow9323

Calculating Favorable Cases (where each person gets down at a different floor)

Options for Person A = Number of floors that the elevator stops at = 6
Options for Person B = 5 and so on

So total number of ways in which the 6 people can each get down at a different floor = 6*5*4*3*2*1 = 6!

To give you additional practice for this logic, here's a follow-up question:

Three people are on an elevator that goes to 6 floors. What is the probability that each person will get out of the elevator at a different floor?

Best Regards

Japinder
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]  21 May 2015, 06:10
1
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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!

Dear shallow9323

Calculating Favorable Cases (where each person gets down at a different floor)

Options for Person A = Number of floors that the elevator stops at = 6
Options for Person B = 5 and so on

So total number of ways in which the 6 people can each get down at a different floor = 6*5*4*3*2*1 = 6!

To give you additional practice for this logic, here's a follow-up question:

Three people are on an elevator that goes to 6 floors. What is the probability that each person will get out of the elevator at a different floor?

Best Regards

Japinder

Thank you.

The practice one is the answer 6*5*4=120 ?
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]  21 May 2015, 06:19
Total no of fav cases-- 6p6

Total no of cases-- 6^6

so ans -- A = 6p6 / 6^6

:D
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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!

Dear shallow9323

Calculating Favorable Cases (where each person gets down at a different floor)

Options for Person A = Number of floors that the elevator stops at = 6
Options for Person B = 5 and so on

So total number of ways in which the 6 people can each get down at a different floor = 6*5*4*3*2*1 = 6!

To give you additional practice for this logic, here's a follow-up question:

Three people are on an elevator that goes to 6 floors. What is the probability that each person will get out of the elevator at a different floor?

Best Regards

Japinder

Thank you.

The practice one is the answer 6*5*4=120 ?

Dear shallow9323

Yes, you got the number of favorable cases right

So, the probability of each person getting out at a different floor = $$\frac{(6*5*4)}{(6*6*6)}$$ = $$\frac{5}{9}$$

Best,

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