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Probability Problem_photographer

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Probability Problem_photographer

by sanalnnair » Fri Sep 10, 2010 12:23 am
Hii, Can anyone help me to solve this question... I can't understand the problem.

Source- Previous GMAT question paper


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


The answer given is A. 5
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by sanju09 » Fri Sep 10, 2010 1:28 am
sorry for a repeat post, consider the under one only
Last edited by sanju09 on Fri Sep 10, 2010 1:34 am, edited 1 time in total.
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by sanju09 » Fri Sep 10, 2010 1:29 am
sanalnnair wrote:Hii, Can anyone help me to solve this question... I can't understand the problem.

Source- Previous GMAT question paper


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


The answer given is A. 5

Good! Training seems working, please hide the OA using "Spoiler Icon" to maintain the charm of knowing answer later.


This problem is simple, consider the six fellows like A < B < C < D < E < F, height wise, and the following possibilities could be counted on finger tips

#1

II row: D E F

I row: A B C

#2

II row: C D F

I row: A B E

#3

II row: C E F

I row: A B D

#4

II row: B D F

I row: A C E

#5

II row: B E F

I row: A C D


Important here is to realize that A and E will have a fixed position in all the arrangements owing to the fact that A is the shortest, which cannot be moved to the right or to the second row, and E is the tallest, which again cannot be moved to the left or to the first column, under the given constraints.


[spoiler]A[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by lokesh r » Fri Sep 10, 2010 3:16 am
This problem can be more easily done considering 6 different numbers rather than taking 6 alphabets and assigning weights to each alphabets.

Let us consider numbers 1,2,3,4,5,6

Arrangements are as below.

456
123

356
124

346
125

256
134

246
135

Is there any standard approach to solve this problem, using Permutations and combinations rule?
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by sanju09 » Fri Sep 10, 2010 3:44 am
lokesh r wrote:This problem can be more easily done considering 6 different numbers rather than taking 6 alphabets and assigning weights to each alphabets.

Let us consider numbers 1,2,3,4,5,6

Arrangements are as below.

456
123

356
124

346
125

256
134

246
135

Is there any standard approach to solve this problem, using Permutations and combinations rule?
No, not all counting cases are dealt with P&C tactics.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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