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PS - Permutations & Combinations

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by f2001290 » Mon Jun 11, 2007 2:17 am
Count the number of patterns by which the 7 passengers in an elevator can get off at the three different floors above the ground floor of the building. (We do not care which passengers get off at a stop, only the pattern of how many of them there are.)

21

3^7

7^3

P(7,3)

C(7,3)

I am getting 36 as answer which is not an option. I don't have OA for this question.
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by mschling52 » Mon Jun 11, 2007 7:22 am
I think the answer is B - 3^7. Since there are 7 passengers, and for each one there are three choices.
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Re: PS - Permutations & Combinations

by jayhawk2001 » Mon Jun 11, 2007 11:58 am
Hmm, At each floor, you can have 7 possible groups (i.e. group of 1,
group of 2, etc.)

So, total across 3 floors = 7*7*7 = 7^3 ?
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by mschling52 » Mon Jun 11, 2007 1:08 pm
I think you are right, Jayhawk, since we are not concerned with which passengers they are, just the number on each floor. My answer of 3^7 would give the number of possible arrangments of the passengers b/w floors if we do consider them unique.
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