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 350,000 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:

Out of seven models, all of different heights, five models [#permalink]
11 Jun 2006, 12:23

Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five modesl are to stand in a line from shortest to tallest, and the fourth-tallets and the sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?

Re: Arrangements of cars [#permalink]
11 Jun 2006, 16:40

acfuture wrote:

Out of seven models, all of different heights, five models will be chosen to pose for a photograph. If the five modesl are to stand in a line from shortest to tallest, and the fourth-tallets and the sixth-tallest models cannot be adjacent, how many different arrangements of five models are possible?

C is right
If we only have to choose 5 from 7 so we got 7C5=21
But we have to exculde some ways in which 4 and 6 is adjacent
If we have 4&6 adjacent so we can't have 5. This way, along with 4 and 6, we must collect 3more models from the rest of 4 models. It means we got 4C3=4ways
And the answer must be 21-4=17

Number of ways 5 models can be chosen from 7 models = 7C5 = 21
Number of ways 4 and 6 are together
12346
13467
12467
23467
that is 4 ways
Hence total = 21-4 = 17

Total number of ways 5 models can be chosen from 7 models = 7C5 = 21

4 and 6 are only adjacent when 5 is not included in the group. Therefore, to calculate the number of ways to choose five people when 5 is not included = 6C5 = 6.

However, we have to subtract 2 because there is a group that does not include 5, but also does not include both 4 and 6....(1, 2, 3, 4, 7) or (1, 2, 3, 6, 7).

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

I had an interesting conversation with a friend this morning, and I realized I need to add a last word on the series of posts on my application process. Five key words:...