How many possible outcomes can be found in these situations?
(a) A board of $10$ mathematicians and $6$ physicists is to be selected from a group of $18$ mathematicians and $12$ physicists.
So for (a), I believe I should use variations without repetitions method.
This leads to $\dfrac{18!}{10!8!} = 43,758$ and $\dfrac{12!}{6!6!}= 924$ Total possible outcomes $= 44,679$
and then I just add them both in the end, but my friend says that I'm wrong and that I need to multiply the values instead. Is that correct? But why if it is so?