In how many ways can a 3-officer slate can be chosen?

Question

The 4-member group has to elect officers. In how many ways can they choose a slate of 3 officers?

I tried to answer it. I used permutation without repetition and I got 24 ways. Is this correct? the word slate makes me doubt my answer. I'm wondering if that means I need to use combination formula.

Answer 1

The word slate makes me believe that the positions for the officers are not different. Hence, I would use the combination formula and say the answer is 4 choose 3. That would be 4!/3! and give you an answer of 4 different ways.

You can also think of this question less combinatoricsy. Lets say there are 4 people: A,B,C,D. Then the group of officers you can have is ABC, ABD, ACD, or BCD. Since it does not matter the order in which you choose the officers, ABC=BAC, and there are only 4 possibilities.

Answer 2

If the positions are different (e.g. you have a president, a secretary, and a treasurer), then there are $4 \cdot 3 \cdot 2=24$ possible slates

If the positions are all the same, then there are just $4$ possible slates, since there are $4$ persons and $1$ person won't get on the slate.