PS - Permutations & Combinations

f2001290 · Posted: Mon Jun 11, 2007 2:17 am Post subject: PS - Permutations & Combinations

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.

mschling52 · Posted: Mon Jun 11, 2007 7:22 am Post subject:

I think the answer is B - 3^7. Since there are 7 passengers, and for each one there are three choices.

jayhawk2001

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 ?

mschling52 · Posted: Mon Jun 11, 2007 1:08 pm Post subject:

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.