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

27 Oct 2007, 12:34

9

This post received KUDOS

17

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

54% (03:06) correct
46% (02:02) wrong based on 471 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?

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

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases: Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: \(C^3_3*C^2_4=6\); Choosing either 4th or 6th tallest models and any 4 from 5 models left: \(C^1_2*C^4_5=10\); Choosing neither 4th nor 6th tallest models: \(C^5_5=1\);

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

Show Tags

20 May 2012, 04:19

9

This post received KUDOS

Total no. of ways "5 models can be selected from 7" = 7C5 = 21.

Fourth tallest and Sixth tallest can never stand adjacent to each other. Also, all selected models will be lined up from shortest to tallest. Hence, we have to pull out the no. of ways 4 & 6 stand adjacent to each other from the above total, i.e = 21.

The following will be the way in which 4 & 6 stand close to each other.

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

Show Tags

24 May 2012, 21:05

7

This post received KUDOS

eyunni wrote:

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

Hi,

Total number of ways 5 models can be selected = \(7C5\) = 21 Total number of ways when 4th & 6th models stand together = 4C3 = 4 (4th & 6th are already chosen, 5th can't be chosen, so only 4 models are left from which 3 are chosen)

Total required ways = Ways of selecting all 5 models - no. of ways when 4th & 6th models are together, =21 - 4 = 17

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

Show Tags

18 Feb 2012, 14:09

1

This post received KUDOS

Bunuel wrote:

eyunni wrote:

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

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases: Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: \(C^3_3*C^2_4=6\); Choosing either 4th or 6th tallest models and any 4 from 5 models left: \(C^1_2*C^4_5=10\); Choosing neither 4th nor 6th tallest models: \(C^5_5=1\);

Total: 6+10+1=17.

Answer: C.

Brilliant, had the same idea, but didn't know how to model it with combinations.

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

Show Tags

24 May 2012, 11:12

1

This post was BOOKMARKED

eyunni wrote:

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

You could also solve it in this way -

Since fourth and sixth tallest cannot be adjacent, the only way for both of them to be in the chosen group and not stand adjacent is to have the fifth tallest model stand between them , as the models have to stand height wise.

Therefore, Desired result = Total ways of choosing models - (ways in which both fourth and sixth tallest are selected ) + (ways in which 4, 5, and sixth tallest are selected) In the above, we add the (ways in which 4, 5, and sixth tallest are selected) because while subtracting the (ways in which both fourth and sixth tallest are selected ) , you would also have eliminated one favorable way. So we are compensating by adding it back.

Total ways of choosing models : \(7C5 = 21\)

Ways in which both fourth and sixth tallest are selected = select them both and select the additional 3 models from remaining 5: \(5C3 = 10\)

Ways in which 4, 5, and sixth tallest are selected = Select 4, 5, and 6 and select the additional 2 models from the remaining 4: \(4C2 = 6\)

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

Show Tags

17 Nov 2012, 08:48

eyunni wrote:

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

5 will be chosen to pose. 5 + 4 + 3 + 2 + 1 = 15

2 can't be adjacent

15 + 2 = 17

sorry for my bad English. This is my 1st time doing Math again in 20 years after I got sober for 2 years

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

Show Tags

28 Dec 2012, 07:24

eyunni wrote:

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

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. \(... 4 5 6 ...\) Now, we need to select 2 more models from 4 remaining. \(=\frac{4!}{2!2!}=6\) Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. \(=6 * 1 = 6\)

Case 2: 4 and NOT 6 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 3: 6 and NOT 4 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 4: 4 and 6 are not selected \(=\frac{5!}{5!}=1\)

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

Show Tags

22 Apr 2013, 23:27

eyunni wrote:

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

1. The size of the larger group is 7. So n is 7. 2. 5 models are to be selected from the larger group. So r is 5 3. The number of combinations without constraint is \(7C5 = 21.\) 4. Let us take the combinations that satisfy the opposite of the constraint i.e., models 4 and 6 being next to each other. This can happen in \(4C3\) ways because 4 and 6 are always selected. Therefore n and r reduce by 2. But 5 is never selected. So n reduces by 1 more. Therefore, n is 4 and r is 3. And \(4C3\) ways is \(4\) ways. 5. (3) - (4) = \(21 -4 =17\). This is the number of combinations with the constraint.

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

Show Tags

23 Apr 2013, 00:42

riyazv2 wrote:

Total no. of ways "5 models can be selected from 7" = 7C5 = 21.

Fourth tallest and Sixth tallest can never stand adjacent to each other. Also, all selected models will be lined up from shortest to tallest. Hence, we have to pull out the no. of ways 4 & 6 stand adjacent to each other from the above total, i.e = 21.

The following will be the way in which 4 & 6 stand close to each other.

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

Show Tags

23 Apr 2013, 04:06

mbaiseasy wrote:

eyunni wrote:

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

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. \(... 4 5 6 ...\) Now, we need to select 2 more models from 4 remaining. \(=\frac{4!}{2!2!}=6\) Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. \(=6 * 1 = 6\)

Case 2: 4 and NOT 6 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 3: 6 and NOT 4 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 4: 4 and 6 are not selected \(=\frac{5!}{5!}=1\)

\(=6 + 5 + 5 + 1 = 17\)

Answer: C

Hi,

In case 2 , how a " Not 6 " is selected ? Will you please explain ?
_________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

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

Show Tags

23 Apr 2013, 04:09

Bunuel wrote:

eyunni wrote:

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

Notice that since the fourth-tallest and sixth-tallest models cannot be adjacent and models are to stand in a line from shortest to tallest, then if we choose 4th and 6th we MUST also choose 5th to stand between them.

Also notice that since the models are to stand in a line from shortest to tallest then for any group of 5 models we choose, there will be only one arrangement possible: from shortest to tallest.

Hence we can have 3 cases: Choosing 4th, 5th, and 6th tallest models and any 2 from 4 models left: \(C^3_3*C^2_4=6\); Choosing either 4th or 6th tallest models and any 4 from 5 models left: \(C^1_2*C^4_5=10\); Choosing neither 4th nor 6th tallest models: \(C^5_5=1\);

Total: 6+10+1=17.

Answer: C.

Excellent and clear approach.
_________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

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

Show Tags

05 May 2013, 11:56

My Approach:

Let the models be 1,2,3,4,5,6 and 7.

1. If models 4,5 and 6 are included in the group of 5 models, then 2 models be selected from remaining 4 Models = 4C2 = 6 2. If model 4 is included and model 6 is not included in the group, then 4 models be selected from the remaining 5 Models = 5C4 = 5 3. If model 6 is included and model 4 is not included in the group, then 4 models be selected from the remaining 5 Models = 5C4 = 5 4. If model 4 and model 6 are not included in the group, then 5 models be selected from the remaining 5 Models = 5C5 = 1

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

Show Tags

05 May 2013, 12:03

kabilank87 wrote:

mbaiseasy wrote:

eyunni wrote:

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

Great question! Thank you!

My approach:

Case 1: 4 and 6 are selected...

This means 5 must also be selected. \(... 4 5 6 ...\) Now, we need to select 2 more models from 4 remaining. \(=\frac{4!}{2!2!}=6\) Since the models already have assigned heights then in each grouping possible, there is only 1 arrangement. \(=6 * 1 = 6\)

Case 2: 4 and NOT 6 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 3: 6 and NOT 4 is selected... This means we need to select 4 from the remaining 5. \(=\frac{5!}{4!1!}=5*1=5\)

Case 4: 4 and 6 are not selected \(=\frac{5!}{5!}=1\)

\(=6 + 5 + 5 + 1 = 17\)

Answer: C

Hi,

In case 2 , how a " Not 6 " is selected ? Will you please explain ?

Case 2: If 6 is not included in the group then 4 and 6 cannot stand adjacent . Case 3: if 6 is not included in the group then 4 and 6 cannot stand adjacent.

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

Show Tags

25 Jul 2014, 05:52

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

23 Sep 2015, 01:25

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.
_________________

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...