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At our disposal is a collection of $10$ red, $11$ blue and $12$ yellow fabrics. (each fabric is unique) In how many ways can we choose $4$ different fabrics if we want at least one fabric of each of the three colors?

My solution was since the first fabric chosen must be red, there are $10$ options for it. Then the next fabric must be blue, which has $11$ options. The third fabric is yellow, with $12$ options, and the last fabric can be any of the colors, provided that it has not already been chosen, so there are $(9+10+11-3)= 30$ ways to choose the last one, making the total number of choices $9\cdot 10\cdot 11\cdot 30$.

My professor said that I needed to divide that by $2$ to get the right answer, but I just don't understand why. Any help would be much appreciated!

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Because under your scheme you would count, for example, both $$R1,B1,Y1,R2\quad\hbox{and}\quad R2,B1,Y1,R1\ .$$ But these are actually the same choice and therefore should not be counted twice.

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The total number of ways is $\binom {10} {2} \binom {11} {1} \binom {12} {1} + \binom {10} {1} \binom {11} {2} \binom {12} {1} + \binom {10} {1} \binom {11} {1} \binom {12} {2} = 19800.$

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