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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?

Position of A and F has to be fix i.t. bottom line left for A and upper line right for F. Other position needs to be filled-in by B, C, D and E.

B can be on right or A or behind A. So 2 positions possible. C can be on right of B or behind him or A, that means 3 possible positions. D and E will have to fit into the positions accordingly without any option. Hence the ans should be 2 X 3 = 6.

Well, to start with I'll at least try to get the answer

Considering the arrangements: 4 5 6 1 2 3

2 4 6 1 3 5

2 5 6 1 3 4

3 4 6 1 2 5

3 5 6 1 2 4

Don't see other arrangements than 5. So would go with (A).

Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!

Well, to start with I'll at least try to get the answer

Considering the arrangements: 4 5 6 1 2 3

2 4 6 1 3 5

2 5 6 1 3 4

3 4 6 1 2 5

3 5 6 1 2 4

Don't see other arrangements than 5. So would go with (A).

Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!

u missed 436 125

_________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Well, to start with I'll at least try to get the answer

Considering the arrangements: 4 5 6 1 2 3

2 4 6 1 3 5

2 5 6 1 3 4

3 4 6 1 2 5

3 5 6 1 2 4

Don't see other arrangements than 5. So would go with (A).

Position of 1 and 6 is fixed. Still don't see a formula for the rest, except manually finding the arrangement, which as you said takes more than 2 mins!

u missed 436 125

But 436 wouldn't be correct as from left to right it has to in increasing order and 4 is greater than 3. Unless I'm not getting it?

The key here is to spot that the tallest (6) and shortest (1) have fixed positions and then count the possible arrangements of others.
_________________

("B" stands for "back", "F" stands for "front", "L" stands for "left", etc.)

Let's also assign "names" to each of the six people - 1 is the shortest, 2 is the next shortest, ... and 6 is the tallest.

Notice first that the only place where 6 can stand is in the BR position. A person standing in any of the other positions has to be shorter than at least one other person, and 6 isn't shorter than anybody.

By similar reasoning, we can see that the only place where 1 can stand is in the FL position. A person standing in any of the other positions has to be taller than at least one other person, and 1 isn't taller than anybody.

So we know that any possible arrangement will be of this form:

BL BM 6 1 FM FR

All we need to do is count possible ways of putting 2, 3, 4, and 5 in positions BL, BM, FM, and FR. In order to count possibilities, let's focus on who goes into the BL position. 1 and 6 are already fixed in their own positions. There's no way 5 could be in the BL position, because there would be no way to assign someone to BM such that the heights in the back row increased consistently from left to right. So we know that the person in the BL position has to be either 2, 3, or 4. We investigate each possibility in turn:

If 2 goes in the BL position, there are just two possibilities:

2 4 6 1 3 5

and

2 5 6 1 3 4

If 3 goes in the BL position, there are also two possibilities:

3 4 6 1 2 5

and

3 5 6 1 2 4

If 4 goes in the BL position, there is just one possible arrangement:

4 5 6 1 2 3

Counting these possibilities, we see that there are only 5 possible arrangements.

After you realize that the tallest and shortest have fixed positions you can just count possible arrangements without any formula.

I immediately figured out that Position 1 and 6 were fixed, but then I ended up wasting my time trying to find a formula for the whole thing. After a while I stopped trying to find a formula and just counted the possible combinations and came up with 5.

Re: A photographer will arrange 6 people of 6 different heights [#permalink]

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12 Jul 2015, 23:45

Here is a way to do this question without using brute force -

Assume the 6 people in ascending order of height are 1,2,3,4,5,6.

Let the positions be

A B C D E F

Now, it is obvious that we have to fix 1 and 6 in the correct positions as shown below:

_ _ 6 1 _ _

Case 1:

Position A = 2

In that case, position B can only be 4 or 5, because, if B is 3, then both 4 and 5 will be greater than 3 (so, none can take positions E and F), and there won't be any case possible.

2 4 6 1 3 5

or

2 5 6 1 3 4

So, 2 solutions.

Case 2:

Position A = 3

In that case, position B cannot be 2. hence only 4 and 5 can take its place.

3 4 6 1 2 5

or

3 5 6 1 2 4

So, 2 solutions.

Case 3:

Position A = 4

In that case, position B can only be 5, and all other positions are also fixed.

4 5 6 1 2 3

So, 1 solution.

Hence, total solutions = 2 + 2 + 1 = 5

This can be done quite fast, but for the purpose of explaining, I have tried to be more elaborate.

In Quant questions such as these, the number of possible arrangements is sometimes so limited that you can actually just list them out - in that way, you can visualize the solution and just do a bit of 'brute force' work:

Here, we're told to arrange 6 people (who all have DIFFERENT heights) into two rows of 3 so that each person in the first row is 'in front' of a taller person in the second row AND heights increase from 'left to right.' We're asked how many arrangements of people are possible.

If we label the six people as 1, 2, 3, 4, 5, 6 (with 1 the shortest, and each number being 'taller' than the one immediately preceding it), we would have the following options...

After you realize that the tallest and shortest have fixed positions you can just count possible arrangements without any formula.

I immediately figured out that Position 1 and 6 were fixed, but then I ended up wasting my time trying to find a formula for the whole thing. After a while I stopped trying to find a formula and just counted the possible combinations and came up with 5.

In addition to realising that Person1 (shortest) has to be in front and Person6 (tallest) has to be in the back row, you also need to realize that there is only one way of arranging the three people in the front and the three people in the back. If the front row has "Person1, Person2 and Person5", this is exactly how they will stand since they must be in increasing order of height from left to right.

Now you have 4 people left (2, 3, 4,5) and you have to choose two of them for the front row. You can do it in 4C2 = 6 ways. But note that you cannot choose both taller people (Person4 and Person5) for the front row because then you will have two shorter people left for the back row and Person3 will be behind Person4. Rest all cases are fine.

Re: A photographer will arrange 6 people of 6 different heights [#permalink]

Show Tags

15 Jul 2016, 05:22

kalpeshchopada7 wrote:

Let me try,

The arrangement will be something like this.

_ _ _ _ _ _

Position of A and F has to be fix i.t. bottom line left for A and upper line right for F. Other position needs to be filled-in by B, C, D and E.

B can be on right or A or behind A. So 2 positions possible. C can be on right of B or behind him or A, that means 3 possible positions. D and E will have to fit into the positions accordingly without any option. Hence the ans should be 2 X 3 = 6.

Is that right?

You just got one thing wrong:

There's one position C is behind A so Be can't be behind A in that position. It's very important to keep track of repeated formations.

Re: A photographer will arrange 6 people of 6 different heights [#permalink]

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30 Dec 2016, 15:19

I got this problem wrong when I encountered it during my practice test but looking at the solution and applying some reverse thinking I came up with following explanation which makes it little easier to quickly solve the problem.

Let's say we have 6 heights = {1, 2, 3, 4, 5, 6}

1st arrangement is obvious which is as following:

4 5 6 <- second row 1 2 3 <- first row

Now we need to think about other possible arrangements with given restrictions.

Couple of things we need to note:

1) Since 1 is smallest and 6 is greatest, they CANNOT be placed elsewhere from their current position. Hence their position is fixed and we need to only concern ourselves with position of rest of the 4 numbers i.e. {2, 3, 4, 5}

2) For other possible arrangements, the numbers from first row will have to go into second and from second row to 1 first row.

3) From {2, 3, 4, 5}, you cannot move both {2, 3} to second row (with {4, 5} moved to 1st row) as it will not satisfy the constraints in any arrangement.

4) Now looking at numbers just from second row, let's say if either of them has to placed in 1st row, they can only occupy 3rd position. Hence we have following two case:

Case 1: Placing {4} in third position of the first row, find arrangements of {2, 3, 5} that satisfy the constraints given in the problem.

_ _ 6 1 _ 4

Case 2: Placing {5} in third position of the first row, find arrangements of {2, 3, 4} that satisfy the constraints given in the problem.

_ _ 6 1 _ 5

In each of the two cases above, you will find two arrangements that satisfy constraints.