92.2k views
5 votes
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?

User Dogahe
by
8.0k points

2 Answers

3 votes

Answer:

4896 ways

Explanation:

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.

User Potatopeelings
by
8.4k points
3 votes

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**

User Grady
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories