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]
04 Mar 2005, 14:59

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can someone explain how we get the second part that will be subtracted from C(7,5)? Yes, i know the possible arrangements can be listed manually, but what is the shorter and more accurate path, using combinations?

My take on the problem. The five models out of seven can be choose such that

1) The 4th tallest and 6th tallest are present in the group of five
2) The 4th tallest is present but not the 6th tallest
3) The 6th tallest is present but not the 4th tallest
4) The 4th tallest and the 6th tallest are not present

Case 1:

The other three models can be choose out of remaining five models in 5C3 ways. Once the five models are choose they can be arranged as following

Consider 4th and 6th together as one group, hence total of 4 models (assuming 4th and 6th together as a single model) in 4! ways and these two models can internally be arranged in 2! ways.

Hence the total ways in which the models can be arranged with 4th and 6th always together is 4!*2!

Without considering any specific constraint total arrnagements are 5!. Hence required arrangements where the 4th and 6th models are never adjacent is

[5! - (2!*4!)]

Case 2:

Once the 4th model is choose, and 6th is not to be choosen, the remaining 4 models can be choosen out of 5 models in 5C4 = 5 ways. These five models can be arrnaged in 5! ways (no constraints on these). Hence total combinations = 5 * 5! ways

Case 3: This is similar to case 2 and hence total ways = 5*5!

Case 4: If 4th and 6th models are excluded then the remaining 5 have to be choosen and they can be arranged in 5! ways.

considering all cases the total number of arrangements are

when you consider the 4th and 6th tallest model to be in the groupe, then the 5th tallest has to be in the group as well, because the models has to stand from short to tallest:

_ _ 4 5 6 or _ 4 5 6 _ => 2 vacant positions

...that means you pick 2 from the remaining 4 => 4c2

consider the case where 4 th and 6th models are always adjacent. We are given that the models are arranged as per their height and if we need the 4th and the 6th models to be adjacent that means that the 5th model cannot be chosen. Automatically the 5th model is ruled out. So what remains is a selection from models 1,2and 3 hence 3c3

Hope this explains....

Would love to know the OA though....

Last edited by swath20 on 06 Mar 2005, 15:04, edited 1 time in total.

consider the case where 4 th and gth models are always chosen. We are given that the models are arranged as per their height and if we need the 4th and the 6th models together that means that 5th cannot be chosen. Automatically the 5th model is ruled out. So what remains is a selection from models 1,2and 3 hence 3c3

Hope this explains....

Would love to know the OA though....

as i mentioned above: the order has to be from shortest to tallest and 4 and 6 cannot stand together. when we choose 4 and 6, 5 has to be between them. so 3 are already chosen and the remaining 2 spaces can be chose from the remaining 4.

Christopher, I am sorry but I don't understand why u are saying the 5th model has to chosen. if the 4th and 6th are to be adjacent then 5th cannot be chosen because if the 5th model is chosen the order would be x,x,4th, 5th, 6th.

What I was trying to do was, from the total no of cases I was reducing the cases where 4th and 6th are adjacent, so it necessarily has to be 1,2,3,4 and 6

No of times when 4th and 6th are not together =Total no cases without restrictions - when models 4th and 6th are adjacent. ( And mind you this will happen only when the 5th model is not chosen) 1,2,3,4 and 6
=7c5-3c3

swath20 has the right approach, except he made a small typo. We need to take out the case where 4th and 6th model are adjacent from C(7,5). When these two people are adjacent, that means they both are selected, and 5th is not selected. We need to select three more people to make it five, from the remaining four people.

agree with Honghu
7C5-1*4C3
unlike other problems in this one there would be only one way to place the 4th and 6th model as per the order required.
Hence, 4C3*1
Correct me if my reasoning seems wrong please