To determine the number of ways to assign first, second, and third place to three students out of nine, we need to calculate the number of permutations of 9 items taken 3 at a time. This is represented by the permutation formula [tex]\(P(n, k)\)[/tex], where [tex]\(n\)[/tex] is the total number of items and [tex]\(k\)[/tex] is the number of items to arrange.
The permutation formula is given by:
[tex]\[
P(n, k) = \frac{n!}{(n - k)!}
\][/tex]
In this problem, [tex]\(n = 9\)[/tex] and [tex]\(k = 3\)[/tex]. Plugging these values into the formula, we get:
[tex]\[
P(9, 3) = \frac{9!}{(9 - 3)!} = \frac{9!}{6!}
\][/tex]
Here, [tex]\(9!\)[/tex] (9 factorial) means the product of all positive integers up to 9:
[tex]\[
9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
And [tex]\(6!\)[/tex] (6 factorial) means the product of all positive integers up to 6:
[tex]\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
We can simplify the fraction because [tex]\(9!\)[/tex] contains all the terms of [tex]\(6!\)[/tex] as factors:
[tex]\[
\frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}
\][/tex]
By canceling out the common [tex]\(6 \times 5 \times 4 \times 3 \times 2 \times 1\)[/tex] terms in the numerator and the denominator, we are left with:
[tex]\[
\frac{9 \times 8 \times 7 \times (6!)}{6!} = 9 \times 8 \times 7
\][/tex]
Calculating the remaining product:
[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 the students is:
[tex]\[
504
\][/tex]
