Start with n=3. That means voting for 3 out of 6 candidates to take 3 vacancies.
Number of combinations of 3 out of 6 candidates is (6)(5)(4)/3!=20. So there are 20 different combinations of candidates. The 3 selected candidates can be assigned to the vacancies in 6 different ways. That gives us 6×20=120 different combinations of 3 candidates with vacancies.
Now, n=2. There are (6)(5)/2!=15 ways of selecting 2 out of 6 candidates to take 3 combinations of vacancies.
Let the candidates be c1, c2, ...c6, and the vacancies v1, v2, v3.
The combinations of candidates are:
c1c2 c2c3 c3c4 c4c5 c5c6
c1c3 c2c4 c3c5 c4c6
c1c4 c2c5 c3c6
c1c5 c2c6
c1c6
And vacancies: v1v2 v1v3 v2v3.
Let’s take any pair of candidates, say, c2c5. They can be assigned to, say, vacancies v1v3:
(c2v1) (c5v3) (c2v3) (c5v1). This applies to the other two combinations of vacancies but the same candidates. So for each pair of candidates, there are 3×4=12 ways to assign the vacancies to the candidates. With 15 different pairs of candidates, we have 12×15=180 different ways of voting.
n=1. 6 ways of selecting the candidates and 3 ways of selecting the vacancies. Each candidate can be assigned to any one of the vacancies, 6×3=18.
n=0. Only one way to vote for nobody.
Total ways of voting: 120+180+18+1=319.