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The vertices of a regular pentagon are to be colored using f

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by Max@Math Revolution » Thu Jul 04, 2019 11:39 pm

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[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
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by deloitte247 » Sun Jul 07, 2019 4:09 am
We can select 5 different colors from a total of 6 colors
$$6_{C_5}=\ \frac{n!}{r!\left(n-r\right)!}=\frac{6!}{5!\left(6-5\right)!}$$
$$=\frac{6\cdot5!}{5!\cdot1!}=6$$
Number of ways in which we can use 5 different colors on the vertices of the pentagon = 5!/5
<Pentagon has 5 vertices>
$$=>\frac{5!}{5}=\frac{5\cdot4\cdot3\cdot2\cdot1}{5}=24$$
Since the 5 colors are to be selected from a palette of 6 different colors, then the pentagon's vertices can be colored in (6 * 24) ways = 144 ways

Correct answer = option D
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by swerve » Sun Jul 07, 2019 9:54 am
We can choose 5 colors from a total of 6 colors in 6C5 ways ie 6 ways !!

We can arrange these 5 chose colors in (5-1!) ways ( since the arrangement is similar to the circular arrangement)

Thus the total number of ways: 6C5 x (5-1)! = 144 ways

Hope this help.
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by Max@Math Revolution » Sun Jul 07, 2019 5:08 pm
=>

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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