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Photographer - Permutation & Combination problem

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Photographer - Permutation & Combination problem

by jitendra_mulchandani » Fri Mar 19, 2010 10:38 pm
86. 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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by kstv » Fri Mar 19, 2010 11:14 pm
Let the persons in the order of Ascending height be 1 2 3 4 5 6
2nd ROW D E F
1st ROW A B C where A to F are the six places
A has to taken by 1 - no one shorter to stand behind him/her of to his left.
F has to be taken by 6 . He cannot be in the first row no on shorter than him to stand behind him. In 2nd Row in D&E he will be have someone shorter than hi, to his left.
B - cannot be taken by 5 as he is the tallest now remember 6 is fixed in F and no one to stand behind. The logic is wrong , ignore this part. So no. of ways 3
C - cannot be taken by 2 as the person to his right has to be shorter than him. Only 1 is shorter and his position is fixed. So no of ways 3. So 1st ROW can be arranged in 3*3 - 9 ways.
IMO E
Last edited by kstv on Sat Mar 20, 2010 5:41 am, edited 1 time in total.
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by jitendra_mulchandani » Sat Mar 20, 2010 12:37 am
IMO A

[spoiler]
(1)
D E F
A B C

(2)
B E F
A C D

(3)
B D F
A C E

(4)
C D F
A B E


(5)
C E F
A B D
[/spoiler]

Any other arrangement violates the two rules within the problem.
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by kstv » Sat Mar 20, 2010 5:37 am
Yes because even the second position can be taken by the 2 and 3. So B can be filled in 2 ways.
Similar problem has been discussed before.
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by outreach » Wed Mar 24, 2010 12:10 am
A B C D E F be the 6 people with increasing heights.

A will always stand in the left corner of row 1 and F will always stand in the right corner of row 2.

so, we basically have 4 people ( B C D E) to arrange in 4 seats. 2 in the front and 2 in the back.

2 people from 4 people who will sit in the first row can be selected in 4C2 ways = 6.
but one of combination will be a case when D and E are in the first row. This is not correct . Atleast one of D or E in the must be in second row. hence subtract 1.
hence, answer = 6 - 1 =5
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