Suppose there are $10$ balls in an urn, $4$ blue, $4$ red, and $2$ green. The balls are also numbered $1$ to $10$. Let $B$ $\geq 0$ represent the number of blue balls in the sample , $R \geq 0$ red balls, and $G \geq 0$ green balls. How many ways are there to select an ordered sample of four balls such that only one of $B,R,G$ is zero.
I asked a problem very like this one few days ago link. But this one has different conditions. I tried to follow the same reasoning.
For a sample of size $4$ with the condition that only one can be zero we have this possibilities:
For $(0,3,1)$ we have $4$ sets to draw from red and $3!$ ways to order them. For the green we have $2$ sets to draw from and $4$ to draw it relative to the red ones. So $4 \cdot 6 \cdot 2 \cdot 4=192$.
For $(3,0,1)$ is the same but with blue, so also $192$
For $(3,1,0)$ we have $4$ sets to draw from blue and $3!=6$ possibilities to order them and $4$ sets to draw red and $4$ to draw them relative to the blue. So $4 \cdot 6 \cdot 4 \cdot 4= 384$
For $(1,3,0)$ is the same, also $384$
However I don't know how to do for $(0,2,2),(2,0,2),(2,2,0)$ because after I draw the first two, how many ways can I order the next two?