HARRISON'S 3 FRIENDS

I have more than 3 friends :)

Anyways, we will use complementary counting to solve this problem. There are a total of $3^6$ ways for me to invite my friends over with no restrictions.

• If a friend comes four times and another two times, this is the same as $3 \cdot 2 \cdot \frac{6!}{4! \cdot 2!} = 90$ ways.
  (This is because there are $3$ ways to choose the first friend and $2$ ways to choose the second friend multiplied by the number of ways to arrange them).
  

• If a friend comes four times and the other two come once, this is the same as $3 \cdot \frac{6!}{4!}=90$ ways

• If a friend comes five times and the other comes once, then this equals $3 \cdot 2 \frac{6!}{5!} =36$ ways

• If a friend comes all six times, then this obviously equals $3$

So our total is equivalent to $3^6-90-90-36-3=510 \implies \boxed{3}$