BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

PS - Permutations & Combinations

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 400
Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member
Thanked: 1 times
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.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 105
Joined: Mon Sep 18, 2006 12:34 pm
Location: OH
Thanked: 7 times
GMAT Score:780

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.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member
Location: Silicon valley, California
Thanked: 30 times

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 ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 105
Joined: Mon Sep 18, 2006 12:34 pm
Location: OH
Thanked: 7 times
GMAT Score:780

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.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1