ajk 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

i found this answer in one of the solution books:

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 confirm 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.