Few days ago I asked almost same question about 3 or more and atmost 1 red balls and thanks to very helpful answers I can calculate any kind of combinations using only two different items(balls).
Now I am facing problem with generalizing this approach. E.I. how this formula $$p=\sum_{j=3}^6 P(\text{green$=j$ and red$=0$})+\sum_{j=3}^5 P(\text{green$=j$ and red$=1$}).$$
transform for the case where ${green}\ge3, \text{red}\le1,\text{orange}\le1$ is required.
Calculating "static" values (E.I ${green}=2, red=1, orange=1$) is breeze with multinomial distribution but these cumulative values really trips me off
In my opinion, the key to this problem is to forgo elegance and count, focusing directly on the # of green balls selected.
My answer will be
$$ \frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
Clearly,
$$D = 4^6.$$
I will compute various numerators $N_1, N_2, N_3, N_4$. Then
$$N = N_1 + N_2 + N_3 + N_4.$$
Assume that the colors are represented by G,R,O, and B (for black).
On each line below, the number of ways that a distribution can occur will be counted.
$\underline{\text{3 Greens} : N_1}$
GGGBBB : $~ \binom{6}{3}.$
GGGBBR : $~ \binom{6}{3} \times 3.$
GGGBBO : $~ \binom{6}{3} \times 3.$
GGGBRO : $~ \binom{6}{3} \times 3!.$
$$N_1 = \binom{6}{3} \times 13.$$
$\underline{\text{4 Greens} : N_2}$
GGGGBB : $~ \binom{6}{4}.$
GGGGBR : $~ \binom{6}{4} \times 2.$
GGGGBO : $~ \binom{6}{4} \times 2.$
GGGGRO : $~ \binom{6}{4} \times 2.$
$$N_2 = \binom{6}{4} \times 7.$$
$\underline{\text{5 Greens} : N_3}$
GGGGGB : $~ \binom{6}{5}.$
GGGGGR : $~ \binom{6}{5}.$
GGGGGO : $~ \binom{6}{5}.$
$$N_3 = \binom{6}{5} \times 3.$$
$\underline{\text{6 Greens} : N_4}$
GGGGGG : $~ \binom{6}{6}.$
$$N_4 = \binom{6}{6} \times 1.$$
final answer:
$$\frac{ \left[\binom{6}{3} \times 13\right] ~+~ \left[\binom{6}{4} \times 7\right] ~+~ \left[\binom{6}{5} \times 3\right] ~+~ \left[\binom{6}{6} \times 1\right] } {4^6}. $$