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Re: Jess has nine different statues and chooses three to arrange [#permalink]
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sandy wrote:
Jess has nine different statues and chooses three to arrange in a display. How many arrangements can she make?

Drill 2
Question: 3
Page: 563


Show: :: OA
504


NOTE: The word arrangement tells us that the order of the statues in the display matters.

Take the task of displaying 3 statues and break it into stages.

Stage 1: Select a statue for the first position
There are nine statues from which to choose. So, we can complete stage 1 in 9 ways

Stage 2: Select a statue for the second position
There are 8 statues remaining. So, we can complete stage 2 in 8 ways

Stage 3: Select a statue for the third position
There are 7 statues remaining. So, we can complete stage 3 in 7 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange three statues) in (9)(8)(7) ways (= 504 ways)

Answer: 504

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn it.

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Jess has nine different statues and chooses three to arrange [#permalink]
Jess has nine different statues

They are not identical

chooses three to arrange

Arrange word will almost always be related to permutation

Permutation - The number of ways to arrange a set of items
Combination - The number of ways to combine a set of items

So, we have to arrange 3 out of 9 statues, which can be done in \(^9P_3\) ways = \(9*8*7\) = 504 ways

If the statues were identical, just divide the above answer by 3!
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Re: Jess has nine different statues and chooses three to arrange [#permalink]
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9P3 ways

i.e. 9!/(9-3)!
= 9!/6!
= 9x8x7x6!/ 6!
= 9x8x7
= 504 ways
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