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:

Out of seven models, all of different heights, five models [#permalink]

Show Tags

30 Nov 2007, 07:54

6

This post received KUDOS

14

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

53% (03:13) correct
47% (02:23) wrong based on 326 sessions

HideShow timer Statistics

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?

Ways in which you can sit all models in oder: 7!/5!2! = 21
Then we figure out in how many of those 21 options model 4 and 6 are sitting together.

I usually draw something like this:
Option 1: _ _ _ _ _ Model 4, 6 could take this spaces, leaving last spot for model 7. You can have 3 in which you can arrange the first 3 models in the first 2 spots = 3

Option 2: _ _ _ _ _ Model 4, 6 could also be on the last 2 spots, but there's only one option on this one, because there are only 3 models to fill the first 3 spots = 1

If the five models are to stand in a line from shortest to tallest

I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?

If the five models are to stand in a line from shortest to tallest

I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?

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?

a) 6 b) 11 c) 17 d) 72 e) 210

Please show your steps

Say we have the models numbered 1~7 ( assume, 1 is shortest and 7 is tallest in that order )

1, 2, 3, 4, 5, 6, 7.

Question says, 4th and 6th can't be adjacent, which implies question "implies" 4th and 6th are always selected amongst the 5 models but never sit/stand together.

(4,6)XXX

The remaining three can be selected in 5C3 ways, and (4 and 6 ) can arrange amongst themselves in 2! ways.

Therefore number of ways choosing,5 models so that 4 and 6 are akways included are

5C3*2! = 20

This also includes the number of ways in which(4,6) are together.

Then, the number of ways in which (4,6) will be always together can be computed

If (4,6) occupy any of the two adjacent places, then remaining three places can be occupied by (xxx) in 3! ways.

If the five models are to stand in a line from shortest to tallest

I feel this condition is irrelevant because in the end the problem asks for the number of ways models can be arranged so that bla bla. It got me confused because in the begining I thought models should stand in ascending order on a photograph =) narrowing down possible ways of arranging them to fewer than 9. what is the source of the problem?

I agree this messed part up as well.

7!/5!*2! = 21 ways

12346 No

12467 No

23467 No

13467 No

so 17 ways.

Ok, agreed. They do stand in ascending order. Therefore, 7!/(5!2!)-4 = 17

Now find out the arrangements where 4 and 6 are NOT adjacent or seat together = total arrangements - Seat together

total arrangements = 7C5 = 21 Seat together = If assume that 4 and 6 are selected in the pool already, need to find out other three members. Therefore we have 7-3 = 4 members left. I deduct 3 because we have to neglect 5 also otherwise 4 and 6 can't be adjacent (the five models are to stand in a line from shortest to tallest - here is the significance). Now select 3 people from 4 members in 4C3 ways. Lets form the equation: the arrangements where 4 and 6 are NOT adjacent or stand together = 7C5 - 4C3 = 21 - 4 = 17.
_________________

If You're Not Living On The Edge, You're Taking Up Too Much Space

lets say 1 2 3 4 5 6 7 are models with 1 to 7 from smallest to tallest.

no of ways to slect any 5 models from 7 models =7C5 =21

no of arrangement for any selection would be =1

there for no of arrangement for 21 selection =21

now 4 and 6 should not be adjusant ,so lets subtract the cases in which 4 and 6 are adjusant.

4 and 6 can be adjusant only when 5 is not selected . so we have decided about selecting two models 4 & 6 and not selecting 5 . no of models left (1 2 3 4 5 6 7)- (4,6 )-5 =1 2 3 7 only four models are left for selection so to make a group of 5 models we have to select 3 out of 4 (or drop 1 out of 4) =4C3 (or 4C1)=4 21-4=17

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?

a) 6 b) 11 c) 17 d) 72 e) 210

Soln: Total number of ways of choosing 5 out of 7 models is = 7C5 Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3 Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2

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?

a) 6 b) 11 c) 17 d) 72 e) 210

Soln: Total number of ways of choosing 5 out of 7 models is = 7C5 Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3 Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2

= 7C5 - 5C3 + 4C2 = 21 - 10 + 6 = 17

Ans is B

Could someone explain me the part highlighted in red? Thanks
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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?

a) 6 b) 11 c) 17 d) 72 e) 210

Soln: Total number of ways of choosing 5 out of 7 models is = 7C5 Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3 Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2

= 7C5 - 5C3 + 4C2 = 21 - 10 + 6 = 17

Ans is B

Could someone explain me the part highlighted in red? Thanks

When we choose 4 and 6 and they are not adjacent means that we must choose 5 too (to stand between them). So for this case we must choose 4, 5, and 6 (3C3) and 2 other from 4 left (4C2) = 3C3*4C2=4C2.

This can be solved in another way:

If we choose 4 and 6, we must also choose 5 (to stand between them) =3C3*4C2=4C2=6 We can choose either 4 or 6 = 2*1C1*5C4=10 We can choose neither 4 nor 6 = 5C5=1

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?

a) 6 b) 11 c) 17 d) 72 e) 210

Soln: Total number of ways of choosing 5 out of 7 models is = 7C5 Number of combinations where 4th tallest and 6th tallest height models will be chosen is = 5C3 Of these combinations of 4th tallest and 6th tallest, the combinations in which 4th and 6th will be chosen but will not come together when arranged in increasing order of height is = 4C2

= 7C5 - 5C3 + 4C2 = 21 - 10 + 6 = 17

Ans is B

Could someone explain me the part highlighted in red? Thanks

When we choose 4 and 6 and they are not adjacent means that we must choose 5 too (to stand between them). So for this case we must choose 4, 5, and 6 (3C3) and 2 other from 4 left (4C2) = 3C3*4C2=4C2.

This can be solved in another way:

If we choose 4 and 6, we must also choose 5 (to stand between them) =3C3*4C2=4C2=6 We can choose either 4 or 6 = 2*1C1*5C4=10 We can choose neither 4 nor 6 = 5C5=1

6+10+1=17.

Answer: C (17).

Hope it helps.

Thanks Bunuel... I missed "are to stand in a line from shortest to tallest"... and hence I was expecting arrangements like x4xx6... too!... My bad
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Out of seven models, all of different heights, five models [#permalink]

Show Tags

06 Jan 2014, 19:46

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Out of seven models, all of different heights, five models [#permalink]

Show Tags

07 Jan 2014, 11:48

2

This post received KUDOS

So 7C5 gives us total number of combinations =21

We need to subtract the number of ways that the 4th and 6th can be next to each other.

The 4th and 6th tallest will only be next to each other when the 5th is NOT selected. The number of combinations where the 5th is not selected is 4 - (two not selected from 7, one of them is the 5th, the others can be 1st, 2nd, 3rd and 7th.)

Re: Out of seven models, all of different heights, five models [#permalink]

Show Tags

10 Jul 2015, 18:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...