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Out of seven models, all of different heights, five models [#permalink]
30 Nov 2007, 07:54

6

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00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

55% (03:07) correct
45% (02:20) wrong based on 215 sessions

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

Re: PS: Permutations & Combinations [#permalink]
22 Dec 2007, 09:45

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?

Re: PS: Permutations & Combinations [#permalink]
22 Dec 2007, 16:45

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?

Re: PS: Permutations & Combinations [#permalink]
22 Dec 2007, 17:09

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

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.

Re: PS: Permutations & Combinations [#permalink]
22 Dec 2007, 22:37

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

Re: PS: Permutations & Combinations [#permalink]
14 Aug 2008, 12:15

4

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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

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

Re: PS: Permutations & Combinations [#permalink]
16 Feb 2010, 11:51

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!!

|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: PS: Permutations & Combinations [#permalink]
16 Feb 2010, 12:22

2

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Expert's post

1

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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

Re: PS: Permutations & Combinations [#permalink]
16 Feb 2010, 13:11

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.

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]
06 Jan 2014, 19:46

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Re: Out of seven models, all of different heights, five models [#permalink]
07 Jan 2014, 11:48

2

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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]
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).

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