To determine the number of ways to place 9 different books on a shelf that can hold only 5 books, we can use the concept of permutations since the order in which the books are placed matters.
First, we need to select 5 books out of the available 9, and then we will arrange these 5 books on the shelf.
-
Selecting the books: We choose 5 books from the 9. The number of ways to choose 5 books from 9 can be calculated using the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n - r)!} \] where \( n \) is the total number of books (9), and \( r \) is the number of books to choose (5). So, \[ \binom{9}{5} = \frac{9!}{5!(9 - 5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126. \]
-
Arranging the selected books: Once we have selected the 5 books, we can arrange these 5 books in any order. The number of ways to arrange 5 books is given by \( 5! \): \[ 5! = 120. \]
-
Calculating total arrangements: The total number of ways to place 9 different books on a shelf that can hold 5 is obtained by multiplying the number of ways to choose the books by the number of ways to arrange them: \[ \text{Total ways} = \binom{9}{5} \times 5! = 126 \times 120 = 15120. \]
Thus, the total number of ways to place 9 different books on a shelf with space for only 5 books is \( \boxed{15120} \).
1 answer