M09 #20 : Retired Discussions [Locked] - Page 2
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# M09 #20

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Intern
Joined: 17 Jan 2012
Posts: 16
Followers: 0

Kudos [?]: 5 [0], given: 8

Re: M09 #20 [#permalink]

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23 Jun 2013, 17:12
Answer: A

The way to approach this problem is using the basic permutation/arrangement formula for n times and removing identical items:

No. of ways to arrange 5 items = 5!
No. of ways to arrange R identical items = 3!
No. of ways to arrange B identical items =2!

Hence total number of ways to arrange = 5! / (3! * 2! ) = 10
Re: M09 #20   [#permalink] 23 Jun 2013, 17:12
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# M09 #20

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