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 such arrangements of the 6 people are possible?

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

I got the answer by listing all possible arrangements on paper. But how to calculate it?

6 people need to be arranged in 2 rows of 3 in such a way that height increases from left to right & the people standing behind in the second row are taller than the person standing in front.

Lets assume the heights of the people (in feet) as 4,5,6,7,8,9 (the question states that all of them are of different heights).

Now, lets arrange the shorter guys in the front row --> 6 5 4 (This can be done in 1 way only).

The second row can be arranged in 3! ways because all the guys in the second row are taller than the guys in the front row.

So, in all the number of way they can be arranged is = 1 X 3! = 6 ways.

Hope my answer is correct

i get 24.

let people in increasing order be A, B, C, D, E, F
=> A is the shortest, and E is the tallest.

R:2 _ _ F
R:1 A _ _

A has to be at the begining of the 1st row as the heights must decrease from left to right. F has to be at the end of the 2nd row, for if we placed him anywhere in the 1st row, we wouldnt have a taller person behind him in the 2nd row and also because heights must decrease from left to right.

2nd row 1st person: can be chosen in 6-2=4 ways.
1st row 2nd person: can be chosen in 6-3=3 ways
2nd row 2nd person chosen in 2 ways.
1st row 3rd person chosen in 1 way

total number of arrangments: 4*3*2*1=24 ways.

hence, D

sorry, my solution is wrong:( i should just guess on such questions.

https://www.beatthegmat.com/awful-photog ... 19792.html

Is there a way not to guess the answer ?

I got the same idea with scooby and stuck there. Do you think that there is a way to solve the problem with out listing all possible combination ?

Vemuri wrote:6 people need to be arranged in 2 rows of 3 in such a way that height increases from left to right & the people standing behind in the second row are taller than the person standing in front.

Lets assume the heights of the people (in feet) as 4,5,6,7,8,9 (the question states that all of them are of different heights).

Now, lets arrange the shorter guys in the front row --> 6 5 4 (This can be done in 1 way only).
The second row can be arranged in 3! ways because all the guys in the second row are taller than the guys in the front row.

So, in all the number of way they can be arranged is = 1 X 3! = 6 ways.

Hope my answer is correct

How about

6 4 2
5 3 1 ??

There are two conditions
1. Height has to increase from left to right and 2.height of people in second row greater than height of first row.

6 X X
X X 1 ... 6 and 1 will be fixed ..they can t be placed in any other position... and rest of the 4 are to be arranged now...

2 can only be placed like

6 X 2
X X 1 or

6 X X
X 2 1

5 can only take 2 places

6 5 X
X X X
OR

6 X X
5 X 1

4 and 3 can take any place of the 4 free positions ....Depending on other conditions

Now the question can be solved.... (I hope so )

First, let us assume the height of the people are 1, 2, 3, 4, 5, and 6. Second, we need to recognize that the positions of the tallest person and the shortest person are set in stone. In other words, the tallest guy is last in the back row, and the shortest guy is first in the front row. Illustration is below.

6 _ _
_ _ 1

Last, we can quickly list out the possible arrangements that satisfies the two criteria (height increases from left to right and person in the back is taller than person in front). Operating on the first criteria of increasing height, we know that the total remaining arrangements in the front row are: 32, 42, 52, 43, 53, and 54. We still have two answer choices, so let us trim the fat with the second criteria. We will see that 54 arrangement cannot be possible, because 5 and 4 are the taller than 3 and 2. If we force the 54 (front row arrangement) scenario we would have this arrangement:

6 3 2
5 4 1

The other remaining arrangements are possible. So we have 5 possible arrangements.

Since the number of people to be arranged are 6 and two of them (1 and 6) have fixed positions and other are highly constrained (to be arranged in increasing order of height) therefore it's best to solve this question manually by taking different possibilities into account without forcibly using P&C techniques to solve this question.

This is a question best approached by taking one condition at a time, and equations will make things even worse, I think. (You're never obliged to use algebra: only use it if it simplifies the problem!)

Let's call our six people ABCDEF 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: this means A must go in the back row. Then, since there's nobody taller than A, he has to go on the right. This gives us

_ _ A
_ _ _

Now let's 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!

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
1XX

Place 2:
Case A:
2X6
1XX

Case B:
XX6
12X

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

Case B:
346...356...456
125...124...123

Total options = 5.

The correct answer is A.