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.
I would use the bijection counting rule. We are mapping: Painters to Electricians using a one-to-one relation.
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.