To determine the number of ways that first, second, and third place can be assigned to three out of nine students, we need to calculate the permutations of 9 students taken 3 at a time. This can be represented mathematically as [tex]\( {}_9P_3 \)[/tex].
The formula for permutations is:
[tex]\[
{}_nP_r = \frac{n!}{(n-r)!}
\][/tex]
where [tex]\(n\)[/tex] is the total number of items (students in this case), and [tex]\(r\)[/tex] is the number of items to choose and arrange.
For this problem:
- [tex]\(n = 9\)[/tex] (the total number of students)
- [tex]\(r = 3\)[/tex] (the number of students to receive the prizes)
Substituting [tex]\(n\)[/tex] and [tex]\(r\)[/tex] into the formula, we get:
[tex]\[
{}_9P_3 = \frac{9!}{(9-3)!}
\][/tex]
First, calculate the factorials involved. The factorial of a number [tex]\(n\)[/tex], denoted [tex]\(n!\)[/tex], is the product of all positive integers up to [tex]\(n\)[/tex].
So, [tex]\(9!\)[/tex] (9 factorial) means:
[tex]\[
9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
And [tex]\((9-3)!\)[/tex], which simplifies to [tex]\(6!\)[/tex], is:
[tex]\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
Now, substitute these values into the permutation formula:
[tex]\[
{}_9P_3 = \frac{9!}{6!}
\][/tex]
Since [tex]\(6!\)[/tex] is a common factor in the numerator and denominator, we can cancel out [tex]\(6!\)[/tex]:
[tex]\[
{}_9P_3 = \frac{9 \times 8 \times 7 \times 6!}{6!}
\][/tex]
[tex]\[
{}_9P_3 = 9 \times 8 \times 7
\][/tex]
Multiplying these values together:
[tex]\[
9 \times 8 = 72
\][/tex]
[tex]\[
72 \times 7 = 504
\][/tex]
Therefore, the number of ways to assign first, second, and third place to three out of nine students is [tex]\( \boxed{504} \)[/tex].
