A basketball team has 5 players. How many different teams can be formed from 15 people if position doesn't matter

You've got 15 choices for the first position, 14 for the second, 13 for the third etc. So for a team of 5 you get 15×14×13×12×11 different ways of picking = 360,360.

The mathematical shorthand for that is:

15!10!=360360

Now that's only correct if we care about who is playing in which position. If we only care about who is in the team, we need to divide that by the number of ways of arranging 5 players in 5 positions. That's given by 5×4×3×2×1 or 5! So if we only care about who's in the team, there are

15!/10!×5!=3003

possible combinations

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: