There are 18 offensive players on the hockey team. How many ways can the coach choose a left wing, center, and right wing to start the game?

We have a total of 18 players, and we want to form groups of three, where the order matters, because there are different roles for each player, so this is a permutation problem.

We solve this problem calculating a permutation of 18 choose 3:

P(18,3) = 18! / (18-3)! = 18! / 15! = 18 * 17 * 16 = 4896

So the coach has 4896 ways to choose the left wing, center and right wing.

micahdisu micahdisu · Answer 2 · 2020-05-07T03:09:25+02:00

Answer: There are 4,896 possible ways

Step-by-step explanation: If there are 18 offensive players on the hockey team, and there is a need to make a selection of three players for three different positions, that means each time you choose one player there is an 18 times 17 (18 x 17) other possibilities for the remaining players. If you choose the next player, there would now be a 17 times 16 (17 x 16) other possibilities for the remaining players, hence we need a formula for the arrangement of 3 players to be chosen from a total of 18.

From the first explanation, if we were to make an arrangement to choose all 18 players the permutation would be given as;

P = 18! (18 factorial)

Which is 18 x 17 x 16 x 15 x 14 x... x 1

However the permutation for selecting just 3 out of the 18 players is given as;

P = 18! ÷ 15!

And this results in;

P = 18 x 17 x 16

P= 4896

Therefore there are 4,896 possible ways

**Note that 18! divided by 15! leaves you with the possible arrangement for 3 persons which is explained as 18 x 17 x 16**