To determine how many ways first, second, and third place can be assigned to three out of nine students, we need to calculate the number of permutations of 9 students taken 3 at a time.
This can be represented mathematically as:
[tex]\[ {}_9 P_3 = \frac{9!}{(9-3)!} \][/tex]
Let's break this down step-by-step:
1. Identify the problem:
Calculate the number of permutations of 9 students taken 3 at a time.
2. Understand the formula for permutations:
Permutations are calculated using the formula:
[tex]\[
{}_n P_r = \frac{n!}{(n-r)!}
\][/tex]
where [tex]\( n \)[/tex] is the total number of items, and [tex]\( r \)[/tex] is the number of items to be chosen.
3. Substitute the values:
For this problem, [tex]\( n = 9 \)[/tex] and [tex]\( r = 3 \)[/tex]. So, substituting these values into the permutations formula gives us:
[tex]\[
{}_9 P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!}
\][/tex]
4. Simplify the factorials:
[tex]\[
9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
[tex]\[
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1
\][/tex]
Plugging these into the equation:
[tex]\[
{}_9 P_3 = \frac{9 \times 8 \times 7 \times 6!}{6!}
\][/tex]
5. Cancel out the common factorials:
Since [tex]\( 6! \)[/tex] appears in both the numerator and the denominator, they cancel each other out:
[tex]\[
{}_9 P_3 = 9 \times 8 \times 7
\][/tex]
6. Multiply the remaining numbers:
[tex]\[
9 \times 8 \times 7 = 504
\][/tex]
Thus, there are 504 ways to assign first, second, and third place to three out of nine students.
