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## Difficult Math Problem #70 - Permutations

tagged by: Brent@GMATPrepNow

This topic has 1 expert reply and 3 member replies
800guy Master | Next Rank: 500 Posts
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#### Difficult Math Problem #70 - Permutations

Mon Dec 04, 2006 2:43 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!

B) 20!/5(4!)

C) 20!/(4!)^5

D) 20!

E) 5!

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gamemaster Senior | Next Rank: 100 Posts
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Tue Dec 05, 2006 2:37 pm
straight forward C

800guy Master | Next Rank: 500 Posts
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Wed Dec 06, 2006 7:34 pm
OA:

4 copies each of 5 types.

Total = 20 books.
Total ways to arrange = 20!

Taking out repeat combos = 20!/(4! * 4! * 4! * 4! * 4!) - each book will have 4 copies that are duplicate. So we have to divide 20! By the repeated copies.

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Roland2rule Master | Next Rank: 500 Posts
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Sat Sep 30, 2017 8:54 pm
I got option E as the correct answer, this is how i solved it. is it correct please?

Using the permutation formula describe below
nPr = n!/ (n-r)!
5P4 = 5!/ (5-4)1 = 5!/1!
= 5!

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Brent@GMATPrepNow GMAT Instructor
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Sun Oct 01, 2017 7:33 am
800guy wrote:
There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!
B) 20!/5(4!)
C) 20!/(4!)^5
D) 20!
E) 5!
Let A, B, C, D, and E represent the 5 different books
So, we want to arrange the following 20 letters: AAAABBBBCCCCDDDDEEEE

-------ASIDE---------------------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

----------ONTO THE QUESTION--------------------------

GIVEN: AAAABBBBCCCCDDDDEEEE
There are 20 letters in total
There are 4 identical A's
There are 4 identical B's
There are 4 identical C's
There are 4 identical D's
There are 4 identical E's
So, the total number of possible arrangements = 20!/[(4!)(4!)(4!)(4!)(4!)]
= 20!/[(4!)^5]

Cheers,
Brent

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