In how many ways can 4 books be arranged on a shelf if there are 8 books to choose from?

Note: [tex]${}_n P_r = \frac{n!}{(n-r)!}$[/tex]

[?] ways



Answer :

Given the problem, we need to determine the number of ways to arrange 4 books out of 8 on a shelf. This is a permutation problem because the order in which the books are arranged matters.

The formula to calculate permutations is given by:

[tex]\[ P(n, r) = \frac{n!}{(n - r)!} \][/tex]

Where:
- \( n \) is the total number of items to choose from,
- \( r \) is the number of items to arrange.

In this problem:
- \( n = 8 \) (the total number of books),
- \( r = 4 \) (the number of books to be arranged).

Using the formula, we substitute the values:

[tex]\[ P(8, 4) = \frac{8!}{(8 - 4)!} \][/tex]
[tex]\[ P(8, 4) = \frac{8!}{4!} \][/tex]

Now, we need to calculate the factorials:

- \( 8! \) (8 factorial) is the product of all positive integers up to 8:
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

- \( 4! \) (4 factorial) is the product of all positive integers up to 4:
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]

Now, divide \( 8! \) by \( 4! \):

[tex]\[ P(8, 4) = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \][/tex]

The \( 4! \) in the denominator cancels out the \( 4! \) in the numerator, which simplifies to:

[tex]\[ P(8, 4) = 8 \times 7 \times 6 \times 5 \][/tex]

Therefore:

[tex]\[ P(8, 4) = 1680 \][/tex]

So, the number of ways to arrange 4 books out of 8 is:

[tex]\[ \boxed{1680} \][/tex]