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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 $10$. 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?

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  • $\begingroup$ Perhaps the $B,R,G $ refer to the balls left in the urn? $\endgroup$
    – RGS
    Oct 27 '17 at 18:34
  • $\begingroup$ @RGS No. Since they're ordered pairs, you get more than one, four in fact. So, $288\cdot4=1152$. Since we have 4 balls. $\endgroup$ Oct 27 '17 at 18:43
  • $\begingroup$ why do you not add also 103, 013 etc. ? $\endgroup$
    – G Cab
    Oct 27 '17 at 18:49
  • $\begingroup$ @GCab Because $103$ and $013$ imply that there are $3$ brown balls. $\endgroup$
    – Numbermind
    Oct 27 '17 at 18:51
  • $\begingroup$ Ah, yes, there are only 2 green balls: sorry $\endgroup$
    – G Cab
    Oct 27 '17 at 21:08
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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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