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A photographer will arrange 6 people of 6 different heights

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

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New post 30 Oct 2009, 07:26
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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?

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

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

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New post 04 Jan 2016, 22:44
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ra011y wrote:
Bunuel wrote:
Merging similar topics.

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.


________ __________ Person6
Person1 _________ _________

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.

So number of ways = 6 - 1 = 5 ways

Answer (A)
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Re: Photo  [#permalink]

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New post 30 Oct 2009, 09:19
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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).
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Re: Photo  [#permalink]

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New post 30 Oct 2009, 09:09
1
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?
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Re: Photo  [#permalink]

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New post 30 Oct 2009, 09:25
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gmattokyo wrote:
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!
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Re: Photo  [#permalink]

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New post 30 Oct 2009, 09:31
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I got 6 as answer too, manually arranging assuming 1,2,3,4,5,6 as heights (since each is different from other)

so possible positions are as below:

456 436
123 125

346 356
125 124

246 256
135 134

Kalpesh logic looks right to me... and is faster than listing possibilities
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Re: Photo  [#permalink]

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New post 30 Oct 2009, 10:32
1
u missed
436
125
:)[/quote]

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?[/quote]


u r right, i got
436
125
incorrectly...

so the answer is 5 ?
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Re: Photo  [#permalink]

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New post 31 Oct 2009, 23:09
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Re: Photo  [#permalink]

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New post 15 Dec 2010, 22:08
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Solution ( courtesy Manhattan GMAT staff)

BL BM BR
FL FM FR

("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.
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Re: GOOD QUESTION  [#permalink]

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New post 24 Feb 2011, 05:11
Bunuel wrote:
Merging similar topics.

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

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New post 12 Jul 2015, 23:45
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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.
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Re: A photographer will arrange 6 people of 6 different heights  [#permalink]

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New post 02 Sep 2015, 21:07
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Hi All,

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...

456
123

356
124

346
125

256
134

246
135

There are NO other options.

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

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New post 04 Jan 2016, 15:23
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Re: A photographer will arrange 6 people of 6 different heights  [#permalink]

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New post 04 Jan 2016, 20:35
Hi kham71,

A number of your examples do NOT follow the 'restrictions' that were mentioned in the prompt. We're told that...

"The heights of the people within each row must INCREASE from left to right...."

So, while 123 would be allowed, 213 would NOT. Keeping that restriction in mind, how many of your possibilities can you eliminate?

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

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New post 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.

Case 1:
1st arrangement:
3 5 6
1 2 4
2nd arrangement:
2 5 6
1 3 4

Case 2:
1st arrangement:
3 4 6
1 2 5
2nd arrangement:
2 4 6
1 3 5


Hence, total arrangements: 1 + 2 + 2 = 5 arrangements.
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Re: A photographer will arrange 6 people of 6 different heights  [#permalink]

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New post 17 Nov 2017, 23:51
Hi Folks,

Just wanted to share my trick to solve this problem.

To answer this choice, you need to know that the positions of 1 and 6 are fixed because they are the lowest and highest.
x y 6
1 v w

v, w, x and y can be any number that create possibility
Therefore, we need to see check the possibility by trial.
The trick here is only v and w that are in play here --> x and y will be automatically fitted from the rule.

Case 1 v,w = 2 and 3
_ _ 6
1 2 3

we got
4 5 6
1 2 3

Case II v,w = 2 and 4
_ _ 6
1 2 4

we got
3 5 6
1 2 4

Case III v,w = 2 and 5
_ _ 6
1 2 5

we got
3 4 6
1 2 5

Case IV v,w = 3 and 4
_ _ 6
1 3 4

we got
2 5 6
1 3 4

Case V v,w = 3 and 5
_ _ 6
1 3 5

we got
2 4 6
1 3 5

Case VI v,w = 4 and 5
_ _ 6
1 4 5

we are left with 2 and 3 which both are less than 4

So, Case VI is no longer valid
Ultimately, we got 5 cases.

Hope this help

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

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New post Updated on: 08 Feb 2018, 19:46
VeritasPrepKarishma wrote:
ra011y wrote:
Bunuel wrote:
Merging similar topics.

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.


________ __________ Person6
Person1 _________ _________

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.

So number of ways = 6 - 1 = 5 ways

Answer (A)


But can't P4 and P5 be arranged in two ways in the front, as P4.P5 or P5.P4. Then, by this method the answer would be (6- 2) = 4 which is incorrect.

What am I missing?

---> stupid qn on my part...can't be P5.P4 from the left because "The heights of the people within each row must increase from left to right"

Originally posted by boston on 06 Feb 2018, 22:23.
Last edited by boston on 08 Feb 2018, 19:46, edited 1 time in total.
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Re: A photographer will arrange 6 people of 6 different heights  [#permalink]

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New post 07 Feb 2018, 00:14
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boston wrote:
But can't P4 and P5 be arranged in two ways in the front, as P4.P5 or P5.P4. Then, by this method the answer would be (6- 2) = 4 which is incorrect.

What am I missing?


4C2 = 6 is the number of ways of selecting 2 people out of 4.
So out of P2, P3, P4 and P5, you can SELECT 2 in 6 ways - (P2, P3), (P2, P4), (P2, P5), (P3, P4), (P3, P5), (P4, P5)

P4, P5 will appear only once in 6.
P4, P5 and P5, P4 is an arrangement but we are only selecting.
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Re: A photographer will arrange 6 people of 6 different heights  [#permalink]

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New post 09 Feb 2018, 10:57
Bunuel wrote:
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


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
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Re: How many such arrangements for photographer?  [#permalink]

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Re: How many such arrangements for photographer?   [#permalink] 01 Feb 2019, 02:46
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