GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Sep 2019, 18:32

### GMAT Club Daily Prep

#### 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

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

# Out of seven models, all of different heights, five models

Author Message
TAGS:

### Hide Tags

VP
Joined: 21 Jul 2006
Posts: 1251
Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

30 Nov 2007, 08:54
6
27
00:00

Difficulty:

95% (hard)

Question Stats:

52% (02:34) correct 48% (02:49) wrong based on 311 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?

A. 6
B. 11
C. 17
D. 72
E. 210
Math Expert
Joined: 02 Sep 2009
Posts: 58117
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

16 Feb 2010, 13:22
5
3
jeeteshsingh wrote:
srivas 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

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.

Hope it helps.
_________________
Senior Manager
Joined: 09 Oct 2007
Posts: 404

### Show Tags

30 Nov 2007, 11:01
6
I got 17, is it correct?

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

21 - (3+1) = 17
##### General Discussion
Senior Manager
Joined: 25 Oct 2006
Posts: 476
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

14 Aug 2008, 13:15
5
1
Here is how I think...

7 models say 1,2,3,4,5,6,7

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
Manager
Joined: 05 Jun 2009
Posts: 73
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

05 Sep 2009, 10:32
4
1
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
Manager
Joined: 27 Oct 2008
Posts: 146
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

27 Sep 2009, 22:31
3
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
Intern
Joined: 02 Oct 2013
Posts: 9
Re: Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

07 Jan 2014, 12:48
2
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.)

So we have 21-4 = 17 ways.
SVP
Joined: 29 Mar 2007
Posts: 2154
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

22 Dec 2007, 17:45
1
1
CaspAreaGuy wrote:
tarek99 wrote:
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.
Director
Joined: 03 Sep 2006
Posts: 689
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

22 Dec 2007, 18:09
1
tarek99 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

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.

Or

46xxx
x46xx
xx46x

(2!*5C3)-3! = 17
Senior Manager
Joined: 03 Apr 2013
Posts: 267
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

16 Jun 2017, 23:41
1
tarek99 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

Okay..here's my take..
Let the models be(in ascending order of height)

A B C D E F G

Now..out of these 7..we have to select 5 and make them stand in ascending order of height.
According to the question, B and D cannot be standing together.

Imagine this, if we select any 5 of these, there will only be one way to make them stand.

Considering the situation, if both B and D were selected, and let's say that B and D were actually selected in the group; then they will stand together every time C is not the part of the group(as there will be no one to stand in the middle). So, our complement event is that B and D are selected, and C is not considered at all for selection. In this way, they will always stand together.

Number of ways to do this..
B and D are already in the group, so we have to select the remaining 3 out of 5. But wait, we also have to never consider C for selection. Finally then, we have to select the remaining 3 out of 4(where C is excluded)

$$4C3 = 4$$

This has to be subtracted from the total possible combinations.

$$7C5 - 4 = 17$$

_________________
Spread some love..Like = +1 Kudos
Manager
Joined: 01 Sep 2007
Posts: 82
Location: Astana
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

22 Dec 2007, 10:45
tarek99 wrote:
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?
Manager
Joined: 01 Sep 2007
Posts: 82
Location: Astana
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

22 Dec 2007, 23:37
GMATBLACKBELT wrote:
CaspAreaGuy wrote:
tarek99 wrote:
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
Manager
Joined: 24 Jul 2009
Posts: 65
Location: United States
GMAT 1: 590 Q48 V24
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

14 Nov 2009, 23:24
kudos srivas, u made it sound simple.......
I was actually planning to post this problem, thank god i searched for it before......
Senior Manager
Joined: 22 Dec 2009
Posts: 264
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

16 Feb 2010, 12:51
srivas 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

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!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~
Senior Manager
Joined: 22 Dec 2009
Posts: 264
Re: PS: Permutations & Combinations  [#permalink]

### Show Tags

16 Feb 2010, 14:11
Bunuel wrote:
jeeteshsingh wrote:
srivas 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

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.

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!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~
Manager
Joined: 14 Jul 2014
Posts: 160
Location: United States
Schools: Duke '20 (D)
GMAT 1: 600 Q48 V27
GMAT 2: 720 Q50 V37
GPA: 3.2
Re: Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

08 Feb 2016, 09:04
Seven models in order: ABCDEFG. D and F should not be adjacent to each other.

Total arrangements: 7C5 = 21.

When D and F are next to each other:
ABCDF
ABDFG
BCDFG
ACDFG

so 21-4 = 17 ways.
Senior Manager
Joined: 04 Aug 2010
Posts: 462
Schools: Dartmouth College
Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

06 Sep 2018, 06:28
tarek99 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

From 7 models, the number of ways to choose 5 to assemble from shortest to tallest = 7C5 = (7*6*5*4*3)/(5*4*3*2*1) = 21.
Among these 21 options, a few will include the 4th and 6th tallest models in adjacent positions and thus will not be acceptable.
The correct answer must be just a bit less than 21.
Eliminate D and E.
Answer choices A and B are so small that they imply that MOST or ALMOST HALF of the 21 options will include the 4th and 6th models in adjacent positions -- not logical.
The only viable answer choice is C.

_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
Intern
Joined: 09 Apr 2018
Posts: 13
Re: Out of seven models, all of different heights, five models  [#permalink]

### Show Tags

22 Mar 2019, 05:04
1
srivas 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

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

Very simple approach ,
Selection of 5 models out of 7 = 7C5
Selection of 5 models such that 4 and 6 are always selected = 5C3
Selection of 5 models excluding 4 and 6 = 7C5-5C3
Selection of 5 models such that 4 and 6 do not stand adjacent, in other words 5 is also selected = 4C2

Different arrangements of five models possible such that 4 and 6 do not stand adjacent = (7C5-5C3) + 4C2
Re: Out of seven models, all of different heights, five models   [#permalink] 22 Mar 2019, 05:04
Display posts from previous: Sort by