Circular Chain

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Circular Chain

by sk8ternite » Wed Aug 26, 2009 1:50 pm
A circular chain consists of 4 gems: a diamond, an emerald, a ruby, and a sapphire. In how many ways can the gems be arranged on the chain?

a. 2
b. 3
c. 6
d. 108
e. 116
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by vani_13in » Wed Aug 26, 2009 3:07 pm
I think (n-1)! i.e. 4-1=3! Ans= 6

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by sk8ternite » Wed Aug 26, 2009 3:20 pm
vani_13in wrote:I think (n-1)! i.e. 4-1=3! Ans= 6
(n-1)!, why did you choose this formula and can you explain the background behind this formula?

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by Stuart@KaplanGMAT » Wed Aug 26, 2009 9:27 pm
sk8ternite wrote:
vani_13in wrote:I think (n-1)! i.e. 4-1=3! Ans= 6
(n-1)!, why did you choose this formula and can you explain the background behind this formula?
That's the general formula for arranging n distinct objects in a circle.

The reason why a circle has fewer permutations than a straight line is because there are n duplicate scenarios for every unique arrangement, depending on where you put the first object.

For example, let's say we're arranging 6 people, A, B, C, D, E and F, around a table. We don't care where they sit, but let's call the chairs 1, 2, 3, 4, 5 and 6.

One arrangement would be:

A1, B2, C3, D4, E5 and F6.

Another arrangement would be:

A2, B3, C4, D5, E6 and F1.

However, looking at the people relative to each other, these two arrangements are identical. So, if we don't factor out n, we have multiple counting.

You could look at the formula as:

n!/n, which is of course the same as (n-1)!
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by praky_rules » Sat Aug 29, 2009 4:53 pm
Answer should n-1!/2 = 3 as the chain can be viewed from either side(front and back!)