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Combinations/Permutations

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Joined: 23 Oct 2011
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Combinations/Permutations [#permalink] New post 30 Nov 2011, 17:08
Combinations/Permutations:

Q1/Q2: I want to Arrange/Select 5 items 2 of which are identical, in 3 places.

Q3/Q4: I want to Arrange/Select 5 items 2 of which are identical, in 7 places.

Any inputs of how to solve this kind of problems?
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Re: Combinations/Permutations [#permalink] New post 01 Dec 2011, 01:57
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Responding to a pm:

Q1: I want to select 3 items out of 5 items (2 of which are identical).
e.g. AABCD
There are 2 different cases:

Case 1: All distinct letters (e.g. ABC, ABD, ACD, BCD)
4C3 i.e. 4 ways

Case 2: 2 letters same (e.g. AAB, AAC, AAD)
Select the two As and any one of the other 3 in 3C1 = 3 ways.

Total = 4+3 = 7 ways

Q2: Arrange 3 letters in 3 places out of 5 letters, 2 of which are identical:
You have already selected the letters above. You further need to arrange them here:

Case 1: Arrange the 3 distinct letters you chose above in 3! ways. So selection and then arrangement gives you
4 * 3! = 4! = 24 ways (e.g. ABC, ACB, BCA, BAC etc)

Case 2: Arrange the 2 identical and 1 different letter selected above in
3 * 3!/2! = 9 ways (e.g. AAB, ABA, BAA etc)

Total = 33 ways

Total = 4+3 = 7 ways
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Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
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Joined: 16 Oct 2010
Posts: 4688
Location: Pune, India
Followers: 1080

Kudos [?]: 4851 [1] , given: 163

Re: Combinations/Permutations [#permalink] New post 01 Dec 2011, 02:08
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SonyGmat wrote:
Combinations/Permutations:

Q3/Q4: I want to Arrange/Select 5 items 2 of which are identical, in 7 places.

Any inputs of how to solve this kind of problems?


Responding to a pm:

You don't need to select here since you anyway have 5 items and more than 5 places. We only need to arrange 5 objects in 7 places. Let's imagine we have two more identical items V. Of the 7 places, wherever we get a vacant spot, we put a V there.

So we have AABCDVV - 7 items (2 of which are imaginary) and 7 places
In how many ways can we arrange 7 items in 7 places?
7! ways.
Since we have 2 identical pairs, we get total number of arrangements = 7!/2!*2!

For more on imaginary place fillers, see:
http://www.veritasprep.com/blog/2011/10 ... ts-part-i/
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Re: Combinations/Permutations   [#permalink] 01 Dec 2011, 02:08
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