# The number of ways of choosing with parameters

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!

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