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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:

$(B,R,G)=\{(0,3,1),(0,2,2),(3,0,1),(2,0,2),(3,1,0),(2,2,0),(1,3,0)\}$

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?

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You correctly calculated the cases in which there are three balls of one color and one ball of another.

Exactly two blue and exactly two red balls: Choose two of the four blue balls and two of the four red balls. Since the four selected balls are numbered, they are all distinct. Hence, they can be arranged in $4!$ orders. Therefore, there are $$\binom{4}{2}\binom{4}{2}4!$$ possible arrangements in this case.

Exactly two blue and exactly two green balls: We must use both of the green balls. Choose two of the four blue balls. As above, there are $4!$ ways to arrange the selected balls. Hence, there are $$\binom{4}{2}4!$$ possible arrangements in this case.

Exactly two red and exactly two green balls: By symmetry, there are also $$\binom{4}{2}4!$$ possible arrangements in this case.

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  • $\begingroup$ In trying to determine where Kyky made an error, I found a simpler way to count the arrangements in which there are two balls of one color and two balls of another color. $\endgroup$ Oct 30 '17 at 23:25

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