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A photographer will arrange 6 people of 6 different heights

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A photographer will arrange 6 people of 6 different heights [#permalink] New post 30 Oct 2009, 06:26
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Question Stats:

54% (02:15) correct 46% (01:28) wrong based on 142 sessions
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
[Reveal] Spoiler: OA

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Re: Photo [#permalink] New post 30 Oct 2009, 08:09
Let me try,

The arrangement will be something like this.

_ _ _
_ _ _

Position of A and F has to be fix i.t. bottom line left for A and upper line right for F. Other position needs to be filled-in by B, C, D and E.

B can be on right or A or behind A. So 2 positions possible. C can be on right of B or behind him or A, that means 3 possible positions. D and E will have to fit into the positions accordingly without any option. Hence the ans should be 2 X 3 = 6.

Is that right?
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Re: Photo [#permalink] New post 30 Oct 2009, 08:19
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Well, to start with I'll at least try to get the answer :)

Considering the arrangements:
4 5 6
1 2 3

2 4 6
1 3 5

2 5 6
1 3 4

3 4 6
1 2 5

3 5 6
1 2 4

Don't see other arrangements than 5. So would go with (A).
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Re: Photo [#permalink] New post 30 Oct 2009, 08:25
gmattokyo wrote:
Well, to start with I'll at least try to get the answer :)

Considering the arrangements:
4 5 6
1 2 3

2 4 6
1 3 5

2 5 6
1 3 4

3 4 6
1 2 5

3 5 6
1 2 4

Don't see other arrangements than 5. So would go with (A).


Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!
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Re: Photo [#permalink] New post 30 Oct 2009, 08:31
I got 6 as answer too, manually arranging assuming 1,2,3,4,5,6 as heights (since each is different from other)

so possible positions are as below:

456 436
123 125

346 356
125 124

246 256
135 134

Kalpesh logic looks right to me... and is faster than listing possibilities
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Re: Photo [#permalink] New post 30 Oct 2009, 08:33
gmattokyo wrote:
gmattokyo wrote:
Well, to start with I'll at least try to get the answer :)

Considering the arrangements:
4 5 6
1 2 3

2 4 6
1 3 5

2 5 6
1 3 4

3 4 6
1 2 5

3 5 6
1 2 4

Don't see other arrangements than 5. So would go with (A).


Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!


u missed
436
125
:)
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Re: Photo [#permalink] New post 30 Oct 2009, 08:41
srini123 wrote:
gmattokyo wrote:
gmattokyo wrote:
Well, to start with I'll at least try to get the answer :)

Considering the arrangements:
4 5 6
1 2 3

2 4 6
1 3 5

2 5 6
1 3 4

3 4 6
1 2 5

3 5 6
1 2 4

Don't see other arrangements than 5. So would go with (A).


Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!


u missed
436
125
:)


But 436 wouldn't be correct as from left to right it has to in increasing order and 4 is greater than 3.
Unless I'm not getting it?
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Re: Photo [#permalink] New post 30 Oct 2009, 09:32
u missed
436
125
:)[/quote]

But 436 wouldn't be correct as from left to right it has to in increasing order and 4 is greater than 3.
Unless I'm not getting it?[/quote]


u r right, i got
436
125
incorrectly...

so the answer is 5 ?
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Re: Photo [#permalink] New post 31 Oct 2009, 22:09
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GOOD QUESTION [#permalink] New post 09 Sep 2010, 15:20
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

is there any algebric methode? I did it by finding all different configuration
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Re: GOOD QUESTION [#permalink] New post 09 Sep 2010, 15:29
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Re: Photo [#permalink] New post 15 Dec 2010, 11:25
Seems to be a very popular problem!
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Re: Photo [#permalink] New post 15 Dec 2010, 21:08
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Solution ( courtesy Manhattan GMAT staff)

BL BM BR
FL FM FR

("B" stands for "back", "F" stands for "front", "L" stands for "left", etc.)

Let's also assign "names" to each of the six people - 1 is the shortest, 2 is the next shortest, ... and 6 is the tallest.

Notice first that the only place where 6 can stand is in the BR position. A person standing in any of the other positions has to be shorter than at least one other person, and 6 isn't shorter than anybody.

By similar reasoning, we can see that the only place where 1 can stand is in the FL position. A person standing in any of the other positions has to be taller than at least one other person, and 1 isn't taller than anybody.

So we know that any possible arrangement will be of this form:

BL BM 6
1 FM FR

All we need to do is count possible ways of putting 2, 3, 4, and 5 in positions BL, BM, FM, and FR. In order to count possibilities, let's focus on who goes into the BL position. 1 and 6 are already fixed in their own positions. There's no way 5 could be in the BL position, because there would be no way to assign someone to BM such that the heights in the back row increased consistently from left to right. So we know that the person in the BL position has to be either 2, 3, or 4. We investigate each possibility in turn:

If 2 goes in the BL position, there are just two possibilities:

2 4 6
1 3 5

and

2 5 6
1 3 4

If 3 goes in the BL position, there are also two possibilities:

3 4 6
1 2 5

and

3 5 6
1 2 4

If 4 goes in the BL position, there is just one possible arrangement:

4 5 6
1 2 3

Counting these possibilities, we see that there are only 5 possible arrangements.
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Re: Photo [#permalink] New post 15 Feb 2011, 12:40
Good question to test logic.. Bunuel is this one of your own?
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Re: GOOD QUESTION [#permalink] New post 24 Feb 2011, 04:11
Bunuel wrote:
Merging similar topics.

After you realize that the tallest and shortest have fixed positions you can just count possible arrangements without any formula.


I immediately figured out that Position 1 and 6 were fixed, but then I ended up wasting my time trying to find a formula for the whole thing. :? After a while I stopped trying to find a formula and just counted the possible combinations and came up with 5.
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Re: A photographer will arrange 6 people of 6 different heights [#permalink] New post 01 Jun 2014, 15:38
Is there any combinatorics approach on this photograph questions? I hope I don't get one of these in the real exam, or should we just rely on brute force and counting outcomes?

Cheers
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Kudos for new approach to this problem :)
Re: A photographer will arrange 6 people of 6 different heights   [#permalink] 01 Jun 2014, 15:38
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