The manager of a sports broadcasting company needs to pick up the best three goals of the month from the available list of ten. How many ways are there to find and decide the top three goals?

This is a problem of combinations. The manager has to choose 3 goals out of 10. The order in which the goals are chosen does not matter, so we use the formula for combinations, which is:

C(n, k) = n! / [k!(n-k)!]

where:

• n is the total number of options,
• k is the number of options to choose,
• "!" denotes factorial, which is the product of all positive integers up to that number.

So in this case, n = 10 (the total number of goals) and k = 3 (the number of goals to choose). Plugging these values into the formula gives:

C(10, 3) = 10! / [3!(10-3)!] = 10! / [3! * 7!] = (10 * 9 * 8) / (3 * 2 * 1) = 720 / 6 = 120

So there are 120 different ways for the manager to choose the top three goals from a list of ten.