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Eight people are to make photo, four of them being standing

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Manager
Joined: 24 Jun 2003
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Eight people are to make photo, four of them being standing [#permalink]  11 Aug 2003, 06:49
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Eight people are to make photo, four of them being standing in the first row and four of them in the second. Every man standing in the front row should be lower that that standing exactly behind him, and all people from the same row should be arranged in order of height increase from left to right. Providing the restrictions mentioned above how many different photos could be taken?
SVP
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No info about their heights. I assume that they all have different heights.
Let their heights be 1, 2, 3, 4, 5, 6, 7, 8 units

Possible variants:

5678
1234

2468
1357

I think 2 variants
Manager
Joined: 10 Jun 2003
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18 (i'll leave out my calcs for now...haha that's assuming i'm correct and they're of any value )
Manager
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stolyar wrote:
No info about their heights. I assume that they all have different heights.
Let their heights be 1, 2, 3, 4, 5, 6, 7, 8 units

Possible variants:

5678
1234

2468
1357

I think 2 variants

Stolyar,

4568
1237

...I think you missed some...
Manager
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the official answer is 14. I can not comment on this.
Let's think together.
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KL

Manager
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THe only way I can think to get 14 is all four in front are the shortest and must line up in one way- Left to right tallest to shortest. Thus in back we are not given the heights, but the tallest has to be the furtherest left. That leaves the other three spots to be filled. Since heights are not given= 4c3+4c2+4c1=14
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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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rich28 wrote:
THe only way I can think to get 14 is all four in front are the shortest and must line up in one way- Left to right tallest to shortest. Thus in back we are not given the heights, but the tallest has to be the furtherest left. That leaves the other three spots to be filled. Since heights are not given= 4c3+4c2+4c1=14

I think your answer is more coincidental than explanatory. Nice try, though.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 70 [0], given: 0

Re: Counting methods # 19 [#permalink]  19 Aug 2003, 00:07
Konstantin Lynov wrote:
Eight people are to make photo, four of them being standing in the first row and four of them in the second. Every man standing in the front row should be lower that that standing exactly behind him, and all people from the same row should be arranged in order of height increase from left to right. Providing the restrictions mentioned above how many different photos could be taken?

I don't know if there is a formula, so I will try to solve this logically:

First, let's say the formation looks like this:

0000
0000

where the heights go from 1 to 8 and the unknowns are 0's.

We know that at a minimum, it must look like this:

0008
1000

Now, consider the person top row 2nd from left. He must have one person shorter than him both to his left and below. So at a minimum, he must be a 4. Similarly, the person in the top row, 3rd from the left must have 5 shorter than him (two on his left, one below plus the two to the left of the one below) so his minimum height is 6.

Now we can count. Let's designate the arrangement as:

XYZ8
1000

Y can only be 4,5, or 6.

If Y is 4, Z can be 6 or 7 and X can be 2 or 3. 4 ways
If Y is 5, Z can be 6 or 7 and X can be 2, 3, or 4. 6 ways.
If Y is 6, Z must be 7 and X can be 2, 3, 4, or 5. 4 ways.

For each of the combinations above, the bottom row is determined (we have 3 numbers left all of which must be in order).

Hence, there are 4 + 6 + 4 = 14 ways.

KL, is there a better way to solve this?
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: Counting methods # 19   [#permalink] 19 Aug 2003, 00:07
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