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How many such arrangements of the 6 people are possible?

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by netigen » Fri May 30, 2008 10:43 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
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by beeparoo » Fri May 30, 2008 12:48 pm
Assuming that A, B, C, D, E, and F represent the varying heights of 6 individuals, wherein A < B < C < D < E <F:

To satisfy the question stem, the shortest and tallest, that is, A & F, must always stand in the following positions:

X X F
A X X

After this, I couldn't figure out how to set up equations, so I just used brute force... Sadly, the unpreferred method.

Arrangement 1:
D E F
A B C

Arrangement 2:
C E F
A B D

Arrangement 3:
C D F
A B E

Arrangement 4:
B E F
A C D

Arrangement 5:
B D F
A C E

Is the answer: 5?
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by netigen » Fri May 30, 2008 2:09 pm
Ans is 5

I am still looking for a better approach ...
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by jasonc » Fri May 30, 2008 2:42 pm
https://nrich.maths.org/public/viewer.ph ... viewer.php
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