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by akash singhal » Thu Nov 12, 2015 9:31 pm
Eight women of eight different heights are to pose for a photo in two rows of four. Each woman in the second
row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row
must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully
adhered to, in how many different ways can the women pose?

(A) 2 (B) 14 (C) 15 (D) 16 (E) 18

OE B

Please Explain How?
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by MartyMurray » Thu Nov 12, 2015 10:32 pm
I am not sure how to do this mathematically. It seems as if that would take quite a formula. So I just wrote them out. Since the women are of eight different heights, I just called them 1, 2, 3, 4, 5, 6, 7, and 8.

5678 4678 4578 4568 3678 3578 3568 3478 3468 2678 2578 2568 2478 2468
1234 1235 1236 1237 1245 1246 1247 1256 1257 1345 1346 1347 1356 1357

Have to stop there, because if you put 14-- in the bottom row, you don't have numbers that work in the top row.

Figuring out how to do this took a while. I very much doubt you would see a question much like this on the actual GMAT. In a way it's good practice though. I had to figure out how to move the numbers around and how to organize the setups.
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by akash singhal » Fri Nov 13, 2015 1:08 am
Marty Murray wrote:I am not sure how to do this mathematically. It seems as if that would take quite a formula. So I just wrote them out. Since the women are of eight different heights, I just called them 1, 2, 3, 4, 5, 6, 7, and 8.

5678 4678 4578 4568 3678 3578 3568 3478 3468 2678 2578 2568 2478 2468
1234 1235 1236 1237 1245 1246 1247 1256 1257 1345 1346 1347 1356 1357

Have to stop there, because if you put 14-- in the bottom row, you don't have numbers that work in the top row.

Figuring out how to do this took a while. I very much doubt you would see a question much like this on the actual GMAT. In a way it's good practice though. I had to figure out how to move the numbers around and how to organize the setups.

Thanks Marty!!
I guess there is no other process to deal with this question.

If any one else has any other solution would be happy to know.
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by GMATGuruNY » Fri Nov 13, 2015 3:48 am
Eight women of eight different heights are to pose for a photo in two rows of four. Each women in the second row must stand directly behind a shorter woman in the first row. In addition, all of the women in each row must be arranged in order of increasing height from left to right. Assuming that these restrictions are fully adhered to, in how many different ways can the women pose?

a. 2
b. 14
c. 15
d. 16
e. 18

OA: B
Let the eight women be the integers 1 through 8, with 1 the shortest and 8 the tallest.

As the SHORTEST, 1 can't stand BEHIND anyone.
As the TALLEST, 8 can't stand IN FRONT OF anyone.
Thus, the positions of 1 and 8 are fixed:
1XXX
XXX8

Case 1: 2 in the front row
12XX
XXX8

A pair of women must stand to the right of 2.
From the 5 remaining women, the total number of pairs that can be formed = 5C2 = 10.
Of these 10 options, one pair -- 67 -- is not viable, since it would force 5 to stand behind 6:
1267
3458

Thus, the total number of VIABLE pairs that can stand to the right of 2 = 10-1 = 9.

Case 2: 2 in the back row, forcing 3 to stand next to 1
13XX
2XX8

A pair of women must stand to the right of 3.
From the 4 remaining women, the total number of pairs that can be formed = 4C2 = 6.
Of these 6 options, one pair -- 67 -- is not viable, since it would force 5 to stand behind 6:
1367
2458

Thus, the total number of VIABLE pairs that can stand to the right of 3 = 6-1 = 5.

Total options = 9+5 = 14.

The correct answer is B.

Here are all of the viable arrangements:

Case 1:
1234...1235...1236...1237...1245...1246...1247...1256...1257
5678...4678...4578...4568...3678...3578...3568...3478...3468

Case 2:
1345...1346...1347...1356...1357
2678...2578...2568...2478...2468
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by [email protected] » Fri Nov 13, 2015 9:31 am
Hi akash singhal,

This prompt is remarkably similar to the one posted here:

https://www.beatthegmat.com/arrangements ... 85387.html

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by Matt@VeritasPrep » Fri Nov 13, 2015 3:01 pm
This question is a pretty close copy of one that appeared on the actual exam: it reminds me of the logic games from the LSAT, but it's fair enough to ask in quant.

Start by realizing that that the tallest woman must go in the back right and the shortest woman must go in the front left.

_ _ _ 1

8 _ _ _

From there, place the second tallest woman. She can go IN FRONT of 1 or to the LEFT of 1, so we have

_ _ 2 1
8 _ _ _

or

_ _ _ 1
8 _ _ 2

Now place the third tallest woman:

_ 3 2 1
8 _ _ _

_ _ 2 1
8 _ _ 3

_ _ 3 1
8 _ _ 2

and continue as such.
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by Matt@VeritasPrep » Fri Nov 13, 2015 3:09 pm
Marty Murray wrote:I am not sure how to do this mathematically. It seems as if that would take quite a formula.
It isn't that bad, I don't think. An approach I like is simply selecting the women who will go in the backline: once you've done this, they'll arrange themselves by height and the front row will also be forced to arrange itself by height.

You can do this by determining whether one or both of women #2 and #3 are in the back. (They can't both be in the front, since they wouldn't both have a taller woman to stand in front of.)

If they BOTH are in the back, your last woman can be any of #4, #5, #6, and #7, so that's four options.

If only #2 is in the back, you can choose any two of (#4, #5, #6, and #7) EXCEPT {#6, #7} (since they wouldn't BOTH have shorter women to stand behind.) So that's (4 choose 2) - 1, or five options.

If only #3 is in the back, same idea: choose any two of (#4, #5, #6, and #7) except {#6, #7}: five options.

Summing this up, we have 5+5+4, or 14.
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