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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.

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  • $\begingroup$ How large is a team? If there are two members to a team, your answer is already OK. $\endgroup$ Nov 25, 2014 at 21:58
  • $\begingroup$ 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. $\endgroup$
    – AdamMc331
    Nov 25, 2014 at 22:00
  • $\begingroup$ But, how large is a team?? $\endgroup$ Nov 25, 2014 at 22:02
  • $\begingroup$ Two members - one painter and one electrician. $\endgroup$
    – AdamMc331
    Nov 25, 2014 at 22:02
  • $\begingroup$ Then there are 8 times 12 = 96 teams $\endgroup$ Nov 25, 2014 at 22:03

1 Answer 1

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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.

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  • $\begingroup$ Why is it not 12!/8!? The answer you gave is a number of about 19.9 million, which seems a bit large $\endgroup$
    – AdamMc331
    Nov 25, 2014 at 21:59
  • $\begingroup$ 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? $\endgroup$
    – ml0105
    Nov 25, 2014 at 22:02
  • $\begingroup$ The logic does make sense. I think we (there are about 5 of us arguing over this) will discuss it with the professor but I do believe that this is a permutation. $\endgroup$
    – AdamMc331
    Nov 25, 2014 at 22:08
  • $\begingroup$ Sounds good! Let us know what the outcome is. :-) $\endgroup$
    – ml0105
    Nov 25, 2014 at 22:10
  • $\begingroup$ Got our quiz back today. We had a really similar question on the quiz, and this was the right answer. I'm so glad I came here first, thank you! $\endgroup$
    – AdamMc331
    Dec 2, 2014 at 21:40

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