Difficult Math Problem #70 - Permutations

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]

Answer: C

Cheers,
Brent

800guy wrote: ↑

Mon Dec 04, 2006 2:43 pm

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!

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