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A company has 8 painters and 12 electricians, and teams can be created of one painter and one electrician. How many different teams can be created?

My best guess is:

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

but I am unsure if that is correct. If it is, am I also correct in assuming the reasoning is because for each of the 8 painters, they could be paired with any of the 12 electricians?

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2 Answers 2

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This is correct. It is a bit simpler to write $8\cdot 12$ though.

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  • $\begingroup$ That was a bit of a brain fart on my side not to notice that. Thanks! $\endgroup$
    – AdamMc331
    Nov 25, 2014 at 16:58
  • $\begingroup$ @McAdam331 no problem. $\endgroup$ Nov 25, 2014 at 16:59
  • $\begingroup$ I had a similar question. More generally, how would you find the number of pairs of people belonging to different groups if there were multiple groups? For example, if there are 4 groups with 3, 5, 2, and 6 people in the groups respectively, how many ways would there be to pick 2 people (each person from two different groups?). I know it's (3*(5+2+6)) + (5*(2+6)) + (2*(6)). How would this be denoted mathematically (I mean, like $ {8 \choose 1} * {12 \choose 1}$) $\endgroup$
    – Shobit
    Feb 22, 2015 at 18:05
  • $\begingroup$ I think the way you wrote it is the best one can do. The moral of the story is that not every problem has a particularly pretty solution. $\endgroup$ Feb 22, 2015 at 18:08
  • $\begingroup$ Ah okay. What I meant was that the solution I wrote was sort of like "brute-force", and I thought there would be some mathematical formula to get to the answer quicker. Thanks, Matt :) $\endgroup$
    – Shobit
    Feb 22, 2015 at 18:20
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Its actually just 12*8, as you can have 12 different electricians and for each you can have 8 different painters. So your answer and guess are perfectly right :)

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