Combinations

swerve wrote: ↑

Sat Feb 06, 2021 6:59 pm

A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller than the person standing in front of him or her. how many such arrangements of the 6 people are possible?

A. 5
B. 6
C. 9
D. 24
E. 36

The OA is A

Source: GMAT Prep

Solution:

We can let the 6 people be A, B, C, D, E, and F, with heights a, b, c, d, e, and f, and a < b < c < d < e < f.

Thus we can have an arrangement that is:

1st row: A, B, C
2nd row: D, E, F

Notice that since A is the shortest, he must be the first (or leftmost) one in the first row. Similarly, since F is the tallest, he must be the last (or the rightmost) one in the second row. Thus the question becomes how many ways to arrange B, C, D and E.

We also see that B can stand in one of the following two positions: either second in the first row or first in the second row. If B is second in the first row, then the arrangements could be either the one listed above or the two listed below:

1st row: A, B, D
2nd row: C, E, F

1st row: A, B, E
2nd row: C, D, F

If B is first in the second row, then the arrangements could be:

1st row: A, C, D
2nd row: B, E, F

1st row: A, C, E
2nd row: B, D, F

Since B can’t stand anywhere else, we have listed all the possible arrangements. We see that there are 5 such arrangements.

Answer: A