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 1207:30 AM PST
-08:30 AM PST
Join the conversation to learn more about INSEAD vs LBS
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 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.
20 Nov 2003, 09:15
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:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
here is the question :
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.
Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)
here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)
than subctracting total minus restricted..
why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.
Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)
here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)
than subctracting total minus restricted..
why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!
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 Data Sufficiency (DS) 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!
20 Nov 2003, 12:03
Kudos
Bookmarks
With this kind of combinatoric problems with limitations the best & the shortest way to solve is to calculate the total number of possible combinations and substract all the cases that fall under the limiting condition.
Specificaly:
a. Total number of combinations is 5C20 = 20!/ (5!15!)
b. Now imagin that you have 5 chairs, 1st, 2nd & 3rd are taken by Pres, VP and accountant respectively. You have only 4th & 5th to be taken by someone else. i.e for the 4th place we have 17 managers, for 5th place we have 16 => C(17,2) = 17!/(2!15!)
c. Total - limits = result
Specificaly:
a. Total number of combinations is 5C20 = 20!/ (5!15!)
b. Now imagin that you have 5 chairs, 1st, 2nd & 3rd are taken by Pres, VP and accountant respectively. You have only 4th & 5th to be taken by someone else. i.e for the 4th place we have 17 managers, for 5th place we have 16 => C(17,2) = 17!/(2!15!)
c. Total - limits = result
22 Nov 2003, 05:32
Kudos
Bookmarks
Leon_v
here is the question :
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.
Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)
here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)
than subctracting total minus restricted..
why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.
Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)
here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)
than subctracting total minus restricted..
why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!
This question is poorly worded. Does this mean that NONE of the 3 can go? or does this mean that any none, one, or two of the 3 can go so long as all 3 don't go?
