<s>[INST] Combinatoric problem with balls 
Suppose there are $10$ balls in an urn, $4$ blue, $4$ red, and $2$ green. The balls are also numbered $1$ to $10$. How many ways are there to select an ordered sample of four balls without replacement such that the number B $\geq 0$  of blue balls, the number $R \geq 0$ of red balls, and the number $G \geq 0$ of green balls are all different?

If $|B|,|R|,|G|$ are all different and  $|B|+|R|+|G|=4$, then there can only be combinations with the numbers $0,1,3$.
So I listed all of the possibilities :
$310$,$130,031,301$
For $310$ we have $3$ blue balls, $1$ red ball and $0$ green. So we have $(4 \times 3 \times 2) \times (4) $ ways
For $130$ we have $1$ blue balls, $3$ red ball and $0$ green. So we have $(4 ) \times (4 \times 3 \times 2) $ ways
For $031$ we have $0$ blue balls, $3$ red ball and $1$ green. So we have $(4 \times 3 \times 2) \times (2)$ ways
For $301$ we have $3$ blue balls, $0$ red ball and $1$ green. So we have $(4 \times 3 \times 2) \times (2)$ ways
Adding up we get $96+96+48+48=288$ 
The real answer should be $1152$. What am I doing wrong?
 [/INST] You've got the color combinations correct: $RBG = [(3,1,0), (1,3,0), (3,0,1), (0,3,1)]$.
For the first color combination, you have $4$ sets of numbers you can draw for the reds, and for those numbers, $6$ orders to draw them in. For the one blue, you have $4$ choices for the number, and $4$ choices for when you draw it relative to the reds. So that's $4 \cdot 6 \cdot 4 \cdot 4 = 384$ ways.
The second color combination is done the same way as the first.
For the third color combination, you have again $4 \cdot 6 = 24$ ways to draw the reds. For the one green, you have $2$ choices for the number, and $4$ choices for when you draw it relative to the reds. So that's $4 \cdot 6 \cdot 2 \cdot 4 = 192$ ways.
The fourth color combination is done the same way as the third.
So we get $384 + 384 + 192 + 192 = 1152$ ways.
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