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Re: A photographer will arrange 6 people of 6 different heights [#permalink]
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12 Mar 2017, 08:14
srini123 wrote: 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 436 125 is not a valid combination. The order should be in ascending order in each row.



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Re: A photographer will arrange 6 people of 6 different heights [#permalink]
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23 Jul 2017, 02:23
So i guess its trail and error for this question.... BTW I got this Q in during my mock test on GMATPrep EP1 Exam 4. This one popped up after I got 6 corrects and ended my first 10 streak T___T
Anyway just watch out if you are planning to do EP1 for mocks.



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Re: A photographer will arrange 6 people of 6 different heights [#permalink]
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28 Jul 2017, 11:04
Sorry i am still not understanding this question. I considered the height of the 6 people to be 1,2,3,4,5,6 and so R2(Row 2) has 4,5,6 and R1 (Row 1) has 1, 2,3 this is the only possible way right because the question also says that the height within the row must be increasing from left to right. Where am i going wrong in my understanding of the question. Thank you



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Re: A photographer will arrange 6 people of 6 different heights [#permalink]
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29 Sep 2017, 20:12
Does this perchance use somewhat of the same logic as the question that ask something along the lines of "how many combinations have Lisa to the right of George"? In those cases, you simply halve the combinatorics problem.
In this case, I figured you could get 6c3 = 20 different groups of three from six people. Then, half of these (10) would have taller people in back. Then, half of those (5) would have each row in the correct order.
Luck, or logic?



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



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Re: A photographer will arrange 6 people of 6 different heights [#permalink]
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03 Feb 2018, 12:35
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 My approach is the following, please, tell me if it is correct Bunuel. We choose in how many ways we can pick 3 people from the 6 people. it is 20. Of this 20, in half the cases the person in the first row will not be taller than the persons in the second row. So ,we are left with 10 cases. Out of this 10, in half the cases the heights will increase from right to left. So, we are left with 5 cases. Is this correct?



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A photographer will arrange 6 people of 6 different heights [#permalink]
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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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07 Feb 2018, 00:14
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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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: A photographer will arrange 6 people of 6 different heights
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