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Math Revolution GMAT Instructor
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
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MathRevolution wrote:
[GMAT math practice question]

The vertices of a regular pentagon are to be colored using five different colors. In how many ways can the pentagon’s vertices be colored if the 5 colors are to be chosen from a palette of 6 different colors?

A. 64
B. 96
C. 108
D. 144
E. 192


Total no of ways pentagon's vertices be colored = 6C5 * 4! = 6*24 = 144

IMO D
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
Combinations/permutations type questions, unless extremely basic, are much harder than sub-600. No chance this is sub-600.
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
Expert Reply
=>

The number of ways to choose \(5\) colors out of \(6\) colors is 6C5 = \(6\).

The number of circular permutations of the \(5\) colors is \((5-1)! = 4! = 24.\)

Thus, \(6*24 = 144\) different colorings of the pentagon are possible.

Therefore, D is the answer.
Answer: D
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
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MathRevolution wrote:
=>

The number of ways to choose \(5\) colors out of \(6\) colors is 6C5 = \(6\).

The number of circular permutations of the \(5\) colors is \((5-1)! = 4! = 24.\)

Thus, \(6*24 = 144\) different colorings of the pentagon are possible.

Therefore, D is the answer.
Answer: D




Hello Mathsrevolution,

Can you please explain the highlighted part

The number of circular permutations of the \(5\) colors is \((5-1)! = 4! = 24.\)
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
Expert Reply
prabsahi wrote:
MathRevolution wrote:
=>

The number of ways to choose \(5\) colors out of \(6\) colors is 6C5 = \(6\).

The number of circular permutations of the \(5\) colors is \((5-1)! = 4! = 24.\)

Thus, \(6*24 = 144\) different colorings of the pentagon are possible.

Therefore, D is the answer.
Answer: D



The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is (n-1)!.
The number is (n-1)! instead of the usual factorial n! since all circular permutations of objects are equivalent because the circle can be rotated.

Hello Mathsrevolution,

Can you please explain the highlighted part

The number of circular permutations of the \(5\) colors is \((5-1)! = 4! = 24.\)



The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is (n-1)!.
The number is (n-1)! instead of the usual factorial n! since all circular permutations of objects are equivalent because the circle can be rotated.
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
How do we know that the order of color doesnot matter in this case?

Suppose that we have green, red, blue, yellow, brown, black-- 6 colors

We can have blue first, then an option of any of the 5 colors in the second wall, then any 1 of the 4 in the 3rd, then any of 3 in the 4th, and then any of the 2 in the 5th.

There are 6 ways to do the same (for each color), therefore answer should be 6!

Why is the above wrong?

Posted from my mobile device
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
MathRevolution wrote:
[GMAT math practice question]

The vertices of a regular pentagon are to be colored using five different colors. In how many ways can the pentagon’s vertices be colored if the 5 colors are to be chosen from a palette of 6 different colors?

A. 64
B. 96
C. 108
D. 144
E. 192

Will we use (n-1)! for triangle too??
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Re: The vertices of a regular pentagon are to be colored using five differ [#permalink]
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