A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created that is made up of 3 sr officers and 1 jr officer, how many different committees are possible?

8
24
58
80*
210

I initially got this one wrong, but after I saw the answer which is 80, I was able to solve it. But im not sure if this is the proper way of solving it, or if someone can suggest a more efficient way of solving it?

This is what I did, which was wrong:
Sr Officers + Jr Officers
(6!/3! 2! = 20) + (4!/3! = 4) = 24

It seems like the correct answer is:
Sr Officers X Jr Officers
(6!/3! 2! = 20) + (4!/3! = 4) = 80

If this is the most efficient way of solving, when do you know when to add them as opposed to multiplying them in this case?

it seems like you're right.

here are the guidelines for that decision:

* if you are selecting both things, independently, then you multiply possibilities.
for instance, a simpler example: if you have to make an outfit from a shirt and a pair of pants, and you have 3 shirts and 4 pairs of pants, that's 3 x 4 = 12 outfits (try listing them and you'll see) - because you have to pick a shirt and a pair of pants.

* if you only select one of the two things - i.e., they are alternatives - then you add possibilities.
if you have to choose either a t-shirt or a polo shirt, and you have 3 t-shirts and 4 polo shirts, that's 7 options.

--

in this case, you're making both decisions, as in the case of the shirt and pants above; they are not alternatives. therefore, you have to multiply the numbers of possibilities. if you could send either three senior officers or one junior officer, then you'd add.