abhi332 wrote: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

[spoiler]OA:B[/spoiler]

Let the eight women be W1,W2,W3.....W8 in ascending order.
Let the arrangement be in
E F G H
A B C D
Now,W1 would always be on 1st A
and W8 will always be on 2nd H
Case 1:-W2 is in B position -W3 can be in C or E,corresponding to it rest can be adjusted in 4 + 6 ways.
Case 2:-W2 is in E position-W3 then in B-W4 can be in F OR C,corresponding to it rest can be adjusted in 6 + 4 ways

Hey abhi are you sure the answer choices are correct??
I am already getting 20 ways??
Or did i do something wrong??

[spoiler]OA:B[/spoiler] is correct

abhi332 wrote:[spoiler]OA:B[/spoiler] is correct

In that case,can you identify where I went wrong??
Can you post the soln. approach??

This ques. also from Manhattan only??

Solution given:

[spoiler]This is a counting problem that is best solved using logic. First, let's represent the line of women as follows:
0000
0000
where the heights go from 1 to 8 in increasing order and the unknowns are designated 0s. Since the women are
arranged by their heights in increasing order from left to right and front to back, we know that at a minimum, the
lineup must conform to this:
0008
1000
Let's further designate the arrangement by labeling the other individuals in the top row as X, Y and Z, and the
individuals in the bottom row as A, B, and C.
XYZ8
1ABC
Note that Z must be greater than at least 5 numbers (X, Y, B, A, and 1) and less than at least 1 number (8). This
means that Z can only be a 6 or a 7.
Note that Y must be greater than at least 3 numbers (X, A and 1) and less than at least 2 numbers (8 and Z). This
means that Y can only be 4, 5, or 6.
Note that X must be greater than at least 1 number (1) and less than at least 3 numbers (8, Z and Y), This means
that X must be 2, 3, 4, or 5.
This is enough information to start counting the total number of possibilities for the top row. It will be easiest to
use the middle unknown value Y as our starting point.
As we determined above, Y can only be 4, 5, or 6. Let's check each case, making our conclusions logically:
If Y is 4, Z has 2 options (6 or 7) and X has 2 options (2 or 3). This yields 2 x 2 = 4 possibilities.
If Y is 5, Z has 2 options (6 or 7), and X has 3 options (2, 3, or 4). This yields 2 x 3 = 6 possibilities.
If Y is 6, Z has 1 option (7), and X has 4 options (2, 3, 4, or 5). This yields 1 x 4 = 4 possibilities.
For each of the possibilities above, the bottom row is completely determined because we have 3 numbers left, all
of which must be in placed increasing order. Hence, there are 4 + 6 + 4 = 14 ways for the women to pose.
The correct answer is (B).
[/spoiler]