The balls are distinguishable and for each ball we have three choices of where to put it. The number of possible ways, without restrictions, is thus $3^7=2187$.
The disallowed ways can be counted by inclusion–exclusion. If the balls are put into two boxes we have three choices for which boxes the balls get put into and two choices per ball, leading to $3×2^7=384$ disallowed ways. We counted the $3×1^7=3$ ways to put all balls into one box twice, so we subtract 3.
Finally, the number of allowed assignments of balls to boxes is $2187-384+3=1806$.
Labeling the colors $A,B,C,D,E,F$ for conciseness, and using an ordered list of sets of colors to represent how the boxes were filled with balls,
consider the arrangement
$$ \{A, B, C\}, \{D\}, \{E\}, \{F\}. $$
You have counted this arrangement six times.
One way you counted it was by making the grouping
$ \{A, B, C\}, \{D\}, \{E\}, \{F\} $
and putting these sets of balls in the boxes in exactly the order they are listed here (i.e., the identity permutation).
Another way you counted it was by making the grouping
$ \{A, B, C\}, \{E\}, \{D\}, \{F\} $
and swapping the middle two sets before putting the sets of balls in the boxes.
The other four ways you counted this arrangement correspond to the other four
permutations of the last three subsets.
Dividing by $6$ won't help, because you counted each of the groupings such as
$ \{A, B\}, \{C, D\}, \{E\}, \{F\} $
exactly four times.
A correct count:
Each of the $\binom{6}{3,1,1,1} = 120$ groupings of the first kind can be distributed into boxes in $\binom 41 = 4$ different ways
(choose which box gets the set of three,
but keep the relative order of equal-sized sets the same).
Each of the $\binom{6}{2,2,1,1} = 180$ groupings of the second kind can be
distributed into boxes in $\binom 42 = 6$ different ways
(choose which boxes get sets of two, etc.).
So that's a total of
$$ 120 \cdot 4 + 180 \cdot 6 = 1560 $$
arrangements.
Note from the comments that $4!{6\brace4} = 24 \cdot 65 = 1560.$
Best Answer
From your examples, the boxes are distinguishable but the balls are not.
Well, since each box has to contain at least one ball, place one ball in each box, leaving you with $N-K$ balls to distribute. The problem now turns into the problem of counting in how many ways can you distribute $N-K$ indistinguishable balls into $K$ distinguishable boxes, with no constraints.
Turns out that it's easier then to simply select the boxes that will have the balls. For your examples, having distributed one ball each in the two boxes, we are left with the problem of placing two balls in two boxes. The first combination corresponds to selecting box number $2$ twice; the second to selecting box number $1$ twice; and the third to selecting box $1$ once and box $2$ once.
So you want to make $N-K$ selections from among $K$ boxes; order does not matter; repetitions are allowed. This is a problem of "combinations with repetitions", also known as the "stars and bars problem". The number of ways of making $s$ selections from among $r$ distinguishable possibilities, where the order does not matter and repetitions are allowed is $$\binom{r+s-1}{s} = \frac{(r+s-1)!}{s!(r-1)!}.$$ In your example, we have $r=K = 2$, and $s=N-K=2$, so $\binom{2+2-1}{2} = \binom{3}{2}=3$ distinct ways.
Plugging in $N-K$ for $s$ and $K$ for $r$, we get $$\binom{K+N-K-1}{N-K} = \binom{N-1}{N-K} = \binom{N-1}{K-1}.$$