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Q25: A photographer will arrange 6 people of 6 different

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Manager
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Joined: 11 Jan 2007
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Q25: A photographer will arrange 6 people of 6 different  [#permalink]

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New post 04 Jul 2007, 20:57
Q25:
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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New post 05 Jul 2007, 05:24
The question sucks! :(

the smallest and tallest persons are fix. so we have just some movement
between the "middle"..
I am not sure - and during a real test I would guess:

4!=4*3*2*1=24

I hope that someone will have a better answer.

Regards.
Mazar
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Re: Probility  [#permalink]

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New post 05 Jul 2007, 05:57
1
jet1445 wrote:
Q25:
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


A, 5.

I called the people 1,2,3,4,5,6.

Another one I just do with brute force. Kind of sucks, but what I realized is that you only need to figure out what can work on the first row. The second row you don't need to count, because whatever's left over must have only one ordering on the second row, because of the constraint that it has to increase.

For example, I know that if 1,2,3 are on the first row, it must be 4,5,6 on the second row.

I soon realized that 1 has to be in the first position, and 6 has to be in the last, so that leaves only 2,3,4,5 to place. If this were anything goes, that would be 4! = 24, but since there are more rules, it has to be less than that. So eliminate D and E.

Now look at row 1. 1 must go first, but if we play with it, we see that 2 and 3 could both go second.

If 2 goes second, 3,4, or 5 can all work in the third spot. (3 orderings)
If 3 goes second, 4 or 5 can both work in the third spot. (2 orderings
If 4 goes second, nothing works, so it can't, and the same for 5.

So in reality, there are only 5 orderings that work.
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Re: Q25: A photographer will arrange 6 people of 6 different  [#permalink]

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New post 07 Mar 2018, 11:30
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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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Re: Q25: A photographer will arrange 6 people of 6 different &nbs [#permalink] 07 Mar 2018, 11:30
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