Awful Photographer

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rabab: Master | Next Rank: 500 Posts; Posts: 132; Joined: Fri Sep 05, 2008 10:19 pm; Location: Bangladesh; Thanked: 1 times

by rabab » Wed Oct 01, 2008 8:06 am

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

Anyone pls? :roll:

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Source: Beat The GMAT — Problem Solving | Wed Oct 01, 2008 8:06 am

Morgoth: Master | Next Rank: 500 Posts; Posts: 316; Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member; Thanked: 36 times

by Morgoth » Wed Oct 01, 2008 9:36 am

IMO 5

A-B-C-D-E-F [arranged in ascending order of height]

first arrangement
DEF - back row
ABC - front row

second arrangement
CEF-back row
ABD - front row

third arrangement
CDF-back row
ABE - front row

fourth arrangement
BEF-back row
ACD - front row

fifth arrangement
BDF-back row
ACE - front row

OA?

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rabab: Master | Next Rank: 500 Posts; Posts: 132; Joined: Fri Sep 05, 2008 10:19 pm; Location: Bangladesh; Thanked: 1 times

by rabab » Wed Oct 01, 2008 9:48 am

Yes OA is A. But could you explain me why are you not considering ABF, CDE and ACF, BDE pairs?

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Morgoth: Master | Next Rank: 500 Posts; Posts: 316; Joined: Mon Sep 22, 2008 12:04 amFollowed by: 1 member; Thanked: 36 times

by Morgoth » Wed Oct 01, 2008 10:06 am

rabab wrote:Yes OA is A. But could you explain me why are you not considering ABF, CDE and ACF, BDE pairs?

"each person in the second row must be taller than the person standing in front of him or her"

"the people in each row must be arranged in ascending order of their height"

The above two statements are most important to the question.

If you consider the two cases

CDE - back row
ABF - front row

This arrangement violates the one of the above two statements.

E has smaller than F therefore E cannot be put behind F

Similarly,

BDE - back row
ACF - front row

E cannot be put behind F since E is smaller than F.

Hope its clear.

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mmslf75: Legendary Member; Posts: 503; Joined: Sun Aug 09, 2009 9:53 pmFollowed by: 2 members; Thanked: 31 times

by mmslf75 » Thu Dec 31, 2009 10:00 am

Morgoth wrote:IMO 5

A-B-C-D-E-F [arranged in ascending order of height]

first arrangement
DEF - back row
ABC - front row

second arrangement
CEF-back row
ABD - front row

third arrangement
CDF-back row
ABE - front row

fourth arrangement
BEF-back row
ACD - front row

fifth arrangement
BDF-back row
ACE - front row

OA?

This method is great, but will take more than 2 mins..THNAKS

any other method

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duyhiep87: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Sun Apr 25, 2010 4:03 pm

by duyhiep87 » Wed Oct 13, 2010 12:22 am

A piece of CAKE!!!!!!!
DE NHU AN CHAO'''
First row: A B C
Second row: D E F

Height: 1 2 3 4 5 6
C must be 6 and D must be 1. Hence, A B 6 and 1 E F.
A B E F must include in {2,3,4,5}
B cannot equal 2 or 3, since if B = 3, E must be 2 and A must be 2. Impossible. Similar when B = 2.
B can be 4 or 5.
If B = 4: A can be 3 or 2; E can be 3 or 2 and F must be 5. We have two solutions of B, A, E, F: (4, 3, 2, 5) (4, 2, 3, 5)
If B = 5: we have 3 solutions of B, A, E, F: (5,4, 2, 3) (5, 3, 2, 4) (5, 2, 3, 4)
Total: 5 solutions. Answer: A

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