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
Is the answer A or E
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i believe it should be A
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Hi sana.noor,
This question hasn't been written in "GMAT style" (the answers haven't been reduced and while it's not stated, I'm going to assume that each person pushes JUST ONE BUTTON).
That having been said, here's a way to think about the math:
1st person = can hit ANY button. = 1/1 This establishes what the other people would need to do.
2nd person = can hit any OTHER button = 5/6
3rd person = can hit any of the 4 OTHERS = 4/6
4th person = can hit any of the 3 OTHERS = 3/6
5th person = can hit any of the 2 OTHERS = 2/6
6th person = can hit the last button = 1/6
1/1 x 5/6 x 4/6 x 3/6 x 2/6 x 1/6 = 5!/(6^5)
This is the same as Answer A
GMAT assassins aren't born, they're made,
Rich
This question hasn't been written in "GMAT style" (the answers haven't been reduced and while it's not stated, I'm going to assume that each person pushes JUST ONE BUTTON).
That having been said, here's a way to think about the math:
1st person = can hit ANY button. = 1/1 This establishes what the other people would need to do.
2nd person = can hit any OTHER button = 5/6
3rd person = can hit any of the 4 OTHERS = 4/6
4th person = can hit any of the 3 OTHERS = 3/6
5th person = can hit any of the 2 OTHERS = 2/6
6th person = can hit the last button = 1/6
1/1 x 5/6 x 4/6 x 3/6 x 2/6 x 1/6 = 5!/(6^5)
This is the same as Answer A
GMAT assassins aren't born, they're made,
Rich
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vipulgoyal
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vipulgoyal
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well here is my solution
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)
SO ON ...........)
So basically # of arrangements of 6 distinct objects: 6!.
P=favorable/total=6!/6^6
Answer: A
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)
SO ON ...........)
So basically # of arrangements of 6 distinct objects: 6!.
P=favorable/total=6!/6^6
Answer: A
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Matt@VeritasPrep
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This is absolutely a GMAT-worthy question, and answers are often not reduced if the form presented is a more natural or elegant one (which is customary in math as a whole).
I think the easiest way to solve this is as follows:
The first person can pick any floor: 6/6
The second can pick any OTHER floor: 5/6
The third can pick any OTHER floor: 4/6
etc.
So we have 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 6!/6^6, as given in a few answers already.
Sana, you asked about answer E: those would be the odds of, say, all six people pushing Floor 3.
I think the easiest way to solve this is as follows:
The first person can pick any floor: 6/6
The second can pick any OTHER floor: 5/6
The third can pick any OTHER floor: 4/6
etc.
So we have 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 6!/6^6, as given in a few answers already.
Sana, you asked about answer E: those would be the odds of, say, all six people pushing Floor 3.
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vipulgoyal
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