Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1207:30 AM PST
-08:30 AM PST
Join the conversation to learn more about INSEAD vs LBS
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. - Dec 16
08:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes! - Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
07 Jan 2004, 01:07
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
100% (00:00) correct
0%
(00:00)
wrong
based on 5
sessions
History
Date
Time
Result
Not Attempted Yet
Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many diffrent ways can the women pose?
(A) 2
(B) 14
(C) 15
(D) 16
(E) 18
please explain
(A) 2
(B) 14
(C) 15
(D) 16
(E) 18
please explain
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
08 Jan 2004, 04:57
Kudos
Bookmarks
In my opinion the answer is A) 2, 1,2,3,4,5,6,7,8 are the women ordered in height from left to right. Fist row 1,2,3,4 second row 5,6,7,8 one way as 5 is behind 1 and so on. Second way first row 1,3,5,7, second row 2,4,6,8, as 2 is behind 1 and so on . The women are ordered in increasing height from left to right and the taller are behind the shorter
09 Jan 2004, 20:44
Kudos
Bookmarks
let the height of each peron be represented as
x+0, x+1, x+2, x+3, x+4, x+5, x+6, x+7
If first row is
x+0,x+1,x+2,x+3,x+4 then 4 people behind them can stand in 4! ways
if first row second row is arranged as
x+0, x+2, x+4, x+6
x+1, x+3,x+5, x+7
here x+1 and x+2 can be exchanged x+3 and x+4 can ge exchanged x+5 and x+6 can be exchanged
These are the only possible ways
so we have 4! + 1 + 3 = 16
x+0, x+1, x+2, x+3, x+4, x+5, x+6, x+7
If first row is
x+0,x+1,x+2,x+3,x+4 then 4 people behind them can stand in 4! ways
if first row second row is arranged as
x+0, x+2, x+4, x+6
x+1, x+3,x+5, x+7
here x+1 and x+2 can be exchanged x+3 and x+4 can ge exchanged x+5 and x+6 can be exchanged
These are the only possible ways
so we have 4! + 1 + 3 = 16
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
