# How many different teams can be created between two groups?

If a company has 8 painters and 12 electricians. How many different teams can be created with 1 painter and 1 electrician?

I know that the number of ways a team can be made is:

${8 \choose 1} * {12 \choose 1}$

Because for each of the eight painters there can be one of the twelve painters, but how many different possibilities of teams are there? I know there will be 8 teams, and that four electricians will sit out, but I don't know how to get the number of possible teams.

• How large is a team? If there are two members to a team, your answer is already OK. Nov 25, 2014 at 21:58
• No, I don't think the above works. The above is just showing how many options for one single team there are. I need to know how many possible teams can exist. Nov 25, 2014 at 22:00
• But, how large is a team?? Nov 25, 2014 at 22:02
• Two members - one painter and one electrician. Nov 25, 2014 at 22:02
• Then there are 8 times 12 = 96 teams Nov 25, 2014 at 22:03

So how many injective functions $f: P \to E$ are there? The answer is $12!/4!$. For $p_{1} \in P$, $f(p_{1})$ has $12$ options. For $p_{2}$, $f(p_{2})$ has $11$ options. We keep proceeding in this manner.
• We are permuting $8$ of the $12$ elements, so we have $P(12, 8) = 12!/(12-8)!$. You can double check the formula here, for example: mathwords.com/p/permutation_formula.htm Combinatorial results often are quite large. So don't be intimidated by that. Does the logic make sense, though? Nov 25, 2014 at 22:02