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

thanks Mitch, very well explained, is there any compact solution i guess it wont fit in 2-3 minut time frame

Starting with
8xxx
Xxx1

We observe that the top row, once filled, completely decides the issue of the bottom row. So this problem is designed to look worse than it is.

The top row can take 6c3 people. That's 20 options.
Of these we can't have 8,x,3,2 (there is no one to sit under 3): minus 4 options.
We also can't have 8,5,4,2 or 8,5,4,3 (no one to sit under 5): minus 2 options.

So the answer is less or equal 14, and for sure greater than 2. Given the answer choices, we can stop here.

misterholmes
www.gmatdojo.com