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!
Difficult Math Problem #70 - Permutations
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Let A, B, C, D, and E represent the 5 different books800guy 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!
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]
Answer: C
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Brent
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We can assume the 20 books are AAAABBBBCCCCDDDDEEEE. Therefore, using the formula for permutation of indistinguishable objects, there are 20!/(4! x 4! x 4! x 4! x 4!) = 20!/(4!)^5 ways to arrange the books on the shelf.
Answer: C
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