# Combinations/Permutations Math Problem: 20 people wish to make up a hockey team.

#### SophieFred316

##### New member
Hello everyone, I am looking for some help with the following math problem.

A dorm floor consisting of 20 people wish to make up a hockey team.

a) In how many ways can you choose 12 people from the 20 people on your dorm floor to make up a hockey team?

* I think for this question I should use combinations - 20C12 = 125,970 ways. However please correct me if I am wrong.

b) For the first name, you must assign 6 out of the 12 people on your team to be forwards, 4 to be defense and 2 to be a goalie. In how many ways can you do this?

Would it be 12C6* 6C4 * 2C2 = 13,860 WAYS?

C) The team decides they want 4 captains with at least one of the captains being a goalie.How many ways can a group of 4 captains with at least one goalie be drawn?

I am not sure how to approach this one.

Thank you!!!!

Last edited:

#### Jomo

##### Elite Member
Hello everyone, I am looking for some help with the following math problem.

A dorm floor consisting of 20 people wish to make up a hockey team.

a) In how many ways can you choose 12 people from the 20 people on your dorm floor to make up a hockey team?

* I think for this question I should use combinations - 20C12 = 125,970 ways. However please correct me if I am wrong.

b) For the first name, you must assign 6 out of the 12 people on your team to be forwards, 4 to be defense and 2 to be a goalie. In how many ways can you do this?

Would it be 12C6* 6C4 * 2C2 = 13,860 WAYS?

C) The team decides they want 4 captains with at least one of the captains being a goalie.How many ways can a group of 4 captains with at least one goalie be drawn?

I am not sure how to approach this one.

Thank you!!!!
Yes, part a and b seem to be correct.

Part c--what does at least one goalie mean in this case? Please answer this and see if you can get somewhere with that. If you can't, then could you answer the question if it said exactly one goalie? If yes, then please show us your work/solution. If not, then what exactly is the trouble?

#### SophieFred316

##### New member
I am not sure if part b should be used in part c?

This is what I did.

2C1*10C3 = 2C2*10C2 = 285 ways