two state event

[Timcago]

A National Hockey League team is allowed to have 18 "skaters" (non-goalies) in uniform for any one league game. Five of these skaters are allowed on the ice at one time. There are three foward positions and two Defence positions on the ice. Only fowards play in the forward positions and only defensemen play in the defence positions. For a particular game, the San Jose Sharks dressed 12 Fowards and 6 Defensemen. How many different sets of five players could Coach Ron Wilson put on the ice at any one time? (hint: consider this a two stage event.) If each set of five players was equally likely to occur, what probability would you assign to each set of players?

I dont know whether to use choose, permutations, or the binomial thing with win lose here.

Can some one help?

 

[pka]

How many different sets of five players could Coach Ron Wilson put on the ice at any one time?
These are clearly combinations:
\(\displaystyle \L {{12} \choose 3}{6 \choose 2}\) is the number of different teams.

Because I know absolutely nothing about any sport, that may be wrong.
I have assumed that all defensive positions are identical as are the forward positions. If they are considered different then use permutations.

 

[galactus]

Because I know absolutely nothing about any sport

Hey pka, a kindred spirit. You and I must be the only two who are not obsessed with who's doing what with what ball. :lol:
In this case, a puck. Though I must admit, palying hockey is fun. I have done that.

 

[Timcago]

Thank you!

12C3=220 sets
6C2=15

So, in total are 220x15=3300 possible sets of players.


Since only one is on the ice, you have 1/3300 probability for a set of players to be on the ice.

Correct?