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by arorag » Mon Sep 01, 2008 11:42 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?
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by sudhir3127 » Mon Sep 01, 2008 11:54 am
arorag 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?
IMO 5.. do let me know if its right so that i can post my solution...
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by arorag » Mon Sep 01, 2008 11:59 am
yes ans is 5
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by sudhir3127 » Mon Sep 01, 2008 12:09 pm
arorag wrote:yes ans is 5
the trick here is to understand the Question properly. the Question stem
says "Every person"

1. Must be shorter than the person standing to their immediate right .

2. Shorter than the one standing directly above them.

thus ,

left most corner guy in the bottom row shud be the shortest and the right most corner guy in the top row shud be the tallest.

now we shud look only @ the unique arrangement of rest of the places..

lets concentrate on the bottom row as if thats done.. top row will have only 1 possible arrangement

the middle postion in the bottom row must be filled in by either 2nd/3th shortest guy

for the second shortest there are 3 options for who can stand to his immediate right

and third shortest will have 2 options

hence 3+2 =5 options,,,

hope that helps. do let me know if u have any doubts..

Can i also know the source of the Question,,, and if u know of any shorter method of solving...
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by crackgmat007 » Thu Jul 30, 2009 2:31 pm
the middle postion in the bottom row must be filled in by either 2nd/3th shortest guy
If all 6 are at different hieghts, IMO there will be only one second shortest guy. Can someone explain if I am missing something?
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by tohellandback » Thu Jul 30, 2009 9:21 pm
did the same question few days ago..only there were 8 women


let the heights in increasing order are 1..to 6

Y1 Y2 Y3
X1 X2 X3

now its clear that X1=1, Y3=6 because no number is greater than 6 and no number is smaller that 1.

Y1 Y2 6
1 X2 X3
Now Y2 must be >1,Y1,X2
and Y2 must be <6
greater than 3 numbers and less than 6. so Y2 must be 4,5

Y1>1
y1<6,Y2
Y1 can be 2,3or 4

so the cases are:
total number of ways:
when Y2=5. y1 can take 3 values
when Y2 =4, y1 can take only 2 values
ans= 2+3=5
The powers of two are bloody impolite!!
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