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
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
First of all, which question are you referring to? There was the original question and then a subsequent question that was modified. The second question was 3 teams, First = 2 people, Second = 2 people and the 3rd = 4 people (total of 8) people.
Can you tell us which question you're talking about so we can explain ourselves?
rino wrote:
I dont see why shouldn't it be a simple 8c2.
i think u are all wrong
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for. _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for.