A library has six identical copies of a certain book. At any given time, some of these copies are at the library and some are checked out. How many different ways are there for some of the books to be in the library and the rest to be checked out if at least one book is in the library and at least one is checked out? (The books should be considered indistinguishable.)
To solve this problem, we can use the concept of stars and bars.
Let's suppose that there are x books in the library, and the remaining 6 - x books are checked out. We need to find the number of ways to distribute these books.
Considering the placement of books as stars and the division between the library and checked out books as bars, we have the following configuration:
|x| |x| |x| |x| |x| |x|
Where '|' represents the bars and 'x' represents the stars.
Now, we need to divide these 6 spaces among the 5 bars. There are (6 - 1) = 5 spaces, and we need to choose 5 - 1 = 4 spaces to place the bars.
Using the stars and bars formula, the number of ways to distribute the books is given by:
C(5, 4) = 5! / (4! * (5 - 4)!) = 5
Therefore, there are 5 different ways to distribute the books between the library and checked out.
To find the number of different ways for some of the books to be in the library and the rest to be checked out, we can break down the problem into cases.
Case 1: 1 book in the library (5 books checked out)
Since we have six identical copies of the book, there are six possible books that could be in the library. Therefore, there are six different ways for one book to be in the library and the rest to be checked out.
Case 2: 2 books in the library (4 books checked out)
To find the number of ways for two books to be in the library, we can choose any two out of the six identical copies. This can be done using the combination formula. Therefore, there are C(6, 2) = 15 different ways for two books to be in the library and the rest to be checked out.
Case 3: 3 books in the library (3 books checked out)
Similarly, we can choose any three out of the six identical copies using the combination formula. Therefore, there are C(6, 3) = 20 different ways for three books to be in the library and the rest to be checked out.
Case 4: 4 books in the library (2 books checked out)
Using the combination formula, we can choose any four out of the six identical copies. Therefore, there are C(6, 4) = 15 different ways for four books to be in the library and the rest to be checked out.
Case 5: 5 books in the library (1 book checked out)
Similarly, there are C(6, 5) = 6 different ways for five books to be in the library and the rest to be checked out.
Case 6: 6 books in the library (0 books checked out)
Lastly, there is only one way for all six books to be in the library and none to be checked out.
Now, we can sum up all the different ways for each case:
6 + 15 + 20 + 15 + 6 + 1 = 63
Therefore, there are 63 different ways for some of the books to be in the library and the rest to be checked out if at least one book is in the library and at least one is checked out.