Final answer:
There are 91 sets of three different numbers that can be chosen from numbers between 0 and 90 with an average of 30.
Step-by-step explanation:
To find the number of sets of three different numbers that can be chosen from numbers between 0 and 90 with an average of 30, we need to consider the range of numbers that satisfy this condition. Let's denote the three numbers as x, y, and z. We know that:
x + y + z = 30 * 3 = 90
Since the numbers have to be different, we need to find the number of ways to choose 3 different numbers that sum up to 90. Let's solve this equation:
90 - x - y = z
Now, we can count the number of pairs (x, y) that satisfy this equation. The possible values for x and y are (0, 90), (1, 89), (2, 88), ..., (90, 0). So, there are 91 possible pairs. Since each pair represents a unique set of three numbers, there are 91 sets of three different numbers with an average of 30.