There were 9 students running in a race. How many different arrangements of first, second, and third place are possible?

To solve this problem, we use the concept of permutations in mathematics.

Step 1: Understand the problem We have 9 students and we want to find out in how many ways we can arrange the first three positions.

Step 2: Apply the formula of permutation The formula for permutation is nPr = n! / (n - r)!, where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

Step 3: Substitute the values into the formula Here, n = 9 (total number of students) and r = 3 (number of positions to fill). So, the permutation is 9P3 = 9! / (9 - 3)!

Step 4: Calculate the factorial 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 and 6! = 6 * 5 * 4 * 3 * 2 * 1

Step 5: Divide 9! by 6! 9P3 = 9! / 6! = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (6 * 5 * 4 * 3 * 2 * 1)

Step 6: Simplify the equation The terms from 6 * 5 * 4 * 3 * 2 * 1 in the numerator and denominator will cancel out, leaving us with 9 * 8 * 7 = 504.

So, there are 504 different arrangements of first, second, and third place possible among 9 students.