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?

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

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.

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

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