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Re: Six Six people are on an elevator that stops at exactly 6 [#permalink]
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]
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]
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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?


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]
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
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Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
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]
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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

Yes, your logic is right.

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]
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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

Yes, your logic is right.

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]
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]
Expert Reply
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

Yes, your logic is right.

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]
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Re: Six people are on an elevator that stops at exactly 6 floors [#permalink]
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