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

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

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New post 30 Nov 2007, 08:54
6
21
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A
B
C
D
E

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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
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Re: PS: Permutations & Combinations  [#permalink]

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New post 16 Feb 2010, 13:22
4
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.

Answer: C (17).

Hope it helps.
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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.
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  [#permalink]

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New post 30 Nov 2007, 11:01
5
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
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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?
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Re: PS: Permutations & Combinations  [#permalink]

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New post 22 Dec 2007, 17:45
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.
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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



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.

Or

46xxx
x46xx
xx46x

Thus answer is

(2!*5C3)-3! = 17
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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
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Re: PS: Permutations & Combinations  [#permalink]

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New post 05 Sep 2009, 10:32
3
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 :-D
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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......
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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
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Re: PS: Permutations & Combinations  [#permalink]

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New post 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.

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

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New post 07 Jan 2014, 12:48
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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.)

So we have 21-4 = 17 ways.
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Re: Out of seven models, all of different heights, five models  [#permalink]

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

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New post 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.

\(7C5 - 4 = 17\)

Answer (C)
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Out of seven models, all of different heights, five models  [#permalink]

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New post 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.


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Out of seven models, all of different heights, five models &nbs [#permalink] 06 Sep 2018, 06:28
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