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# A photographer will arrange 6 people of 6 different heights

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Senior Manager
Joined: 05 Jun 2008
Posts: 289
A photographer will arrange 6 people of 6 different heights [#permalink]

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24 Sep 2008, 07:27
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A photographer will arrange 6 people of 6 different heights for photograph by placing
them in two rows of three so that each person in the first row is standing in front of
someone in the second row. The heights of the people within each row must increase
from left to right, and each person in the second row must be taller than the person
standing in front of him or her. How many such arrangements of the 6 people are
possible?
A. 5
B. 6
C. 9
D. 24
E. 36

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Intern
Joined: 02 Sep 2008
Posts: 45
Re: Combinatory Problem - Help needed [#permalink]

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24 Sep 2008, 07:28

Senior Manager
Joined: 04 Aug 2008
Posts: 363
Re: Combinatory Problem - Help needed [#permalink]

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24 Sep 2008, 09:03
Looks like 36 is the answer if I got it right.
6 people w diff height lets assign them numbers from 1 shortest to 6 tallest

so we have 2 rows

1 2 3
4 5 6

three shortest will always stay in the back and three tallest in front row

therefore back row can be combined 1*2*3 = 6
same for front row

6x6=36
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Intern
Joined: 02 Sep 2008
Posts: 45
Re: Combinatory Problem - Help needed [#permalink]

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25 Sep 2008, 09:42
Solution:

There are 6 people and all have different heights.

The persons are in the back has to be taller than the person in the front.

Let's say a1, a2, a3, a4, a5, a6 are 6 persons.

The persons in ascending order based on height.

a1 < a2 < a3 < a4 < a5 < a6.

Condition: The heights of the people within each row must increase from left to right.

combinations:

1) a4 a5 a6
a1 a2 a3

2) a2 a5 a6
a1 a3 a4

3) a3 a5 a6
a1 a3 a4

4) a2 a4 a6
a1 a3 a5

5) a3 a4 a6
a1 a2 a5

So total five possibilities. Other are ignored based on above conditions.

Senior Manager
Joined: 05 Jun 2008
Posts: 289
Re: Combinatory Problem - Help needed [#permalink]

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25 Sep 2008, 11:01
Twoone wrote:
Solution:

There are 6 people and all have different heights.

The persons are in the back has to be taller than the person in the front.

Let's say a1, a2, a3, a4, a5, a6 are 6 persons.

The persons in ascending order based on height.

a1 < a2 < a3 < a4 < a5 < a6.

Condition: The heights of the people within each row must increase from left to right.

combinations:

1) a4 a5 a6
a1 a2 a3

2) a2 a5 a6
a1 a3 a4

3) a3 a5 a6
a1 a3 a4

4) a2 a4 a6
a1 a3 a5

5) a3 a4 a6
a1 a2 a5

So total five possibilities. Other are ignored based on above conditions.

Is thier any other way solve it, this solution involve lot many steps may not be fisible to attempt at GMAT.
Senior Manager
Joined: 04 Jan 2006
Posts: 276
Re: Combinatory Problem - Help needed [#permalink]

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25 Sep 2008, 11:35
vivektripathi wrote:
Is thier any other way solve it, this solution involve lot many steps may not be fisible to attempt at GMAT.

This is my solution. Just draw diagram and check each possible solution.

Case 1: Basic set up. The smallest guy on lower left and the highest guy on upper right.

4 5 6
1 2 3

You cannot move 6 down or move 1 up because this will violate the rule. Therefore, guy 2, 3, 4, and 5 will be moving around.

Case 2: Move 5 down and 3 up
3 4 6
1 2 5

Case 3: Move 5 down and 2 up
2 4 6
1 2 5

Case 4: Move 4 down and 3 up
3 5 6
1 2 4

Case 5: Move 4 down and 2 up
2 5 6
1 3 4

5 cases total. The best approach to this soltuion should be this way because there are too many conditions and the number of people are not that many.
Intern
Joined: 06 May 2008
Posts: 27
Re: Combinatory Problem - Help needed [#permalink]

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06 May 2009, 13:45
spiridon wrote:
Looks like 36 is the answer if I got it right.
6 people w diff height lets assign them numbers from 1 shortest to 6 tallest

so we have 2 rows

1 2 3
4 5 6

three shortest will always stay in the back and three tallest in front row

therefore back row can be combined 1*2*3 = 6
same for front row

6x6=36

If you combine each row as 6!, you are not considering that the first two guys at each row must have a taller guy at his left. Its not a free combination... it has restrictions.

5 seems to be the answer for me...

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: Combinatory Problem - Help needed   [#permalink] 06 May 2009, 13:45
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