Problem Solving

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 are such arrangements of the 6 people possible?

(A) 5
(B) 6
(C) 9
(D) 24
(E) 36

OA : 5

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 some one in second row. The heights of the people with in each row should be increasing from left 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

Let the 6 people be represented by the numbers 1-6, inclusive, where 1 is the shortest and 6 is the tallest.
Fill the MOST RESTRICTED positions first and work down to the LEAST RESTRICTED positions.

Place 1 and 6:
XX6<---SECOND ROW
1XX<---FIRST ROW

Place 2:
Case A:
2X6<---SECOND ROW
1XX<---FIRST ROW

Case B:
XX6<---SECOND ROW
12X<---FIRST ROW

Place 5, whose position will determine where 3 and 4 can go:
Case A:
256...246<---SECOND ROW
134...135<---FIRST ROW

Case B:
346...356...456<---SECOND ROW
125...124...123<---FIRST ROW

Total options = 5.

The correct answer is A.

Let's elaborate a little bit.

We'll call our six people A, B, C, D, E, and F with heights such that A > B > C > D > E > F.

A is the tallest, so we'll start with him. There's nobody taller than A, so nobody can stand behind A, meaning A must go in the back row. Since there can't be anyone taller than A in that row, he has to go on the right. This gives us

_ _ A
_ _ _

Now consider F. There's nobody shorter than F, so using logic just like we used above, F must go in the front row on the far left.

_ _ A
F _ _

Now we have a few options to play with. Let's start by placing B. Since B is the tallest person left, we have TWO options for him: he can go in front of A, or to the left of A.

_ B A
F _ _

or

_ _ A
F _ B

Now we'll place the shortest remaining person, E. Much like B above, E only has two options: he can go behind F, or to the right of F. Combining this with the above deductions, we now have four arrangements.

E B A
F _ _

_ B A
F E _

E _ A
F _ B

_ _ A
F E B

Now let's look at our four options carefully. If C and D are in the SAME COLUMN, we can't choose their arrangement: C is taller, so he goes on the right. Likewise, if C and D are in the SAME ROW, we can't choose their arrangement: C is taller, so he goes in the back.

In the first, third, and fourth arrangements above, C and D are either in the same column or are in the same row, so those three arrangements are DONE, and we have THREE arrangements so far.

The only arrangement we can play with is the second one, which has two permutations:

C B A
F E D

or

D B A
F E C

So we have a total of FIVE arrangements: the three that were DONE in the previous step, and the two more we just discovered.

Not as bad as it looks, if you take it condition by condition!