Question

In how many ways can 2 red, 2 black, 3 white and 2 blue balls be selected from 4 red, 3 black, 4 white and 8 blue balls? In how many ways can they be arranged?

Answer 1

From the total pool of colored balls, one can choose 2 reds, 2 blacks, 3 whites, and 2 blues in

$\dbinom42\cdot\dbinom32\cdot\dbinom43\cdot\dbinom82=6\cdot3\cdot4\cdot28=2016$

ways.

I'm assuming no ball of the same color is distinguishable from any other ball of the same color. So when I'm considering the possible arrangements, if I had lined up the ball as

red1 - black - red2 - ...

then this would be no different that

red2 - black - red1 - ...

So I now have 9 balls to arrange, which means there are $9!=362,880$ total possible permutations of them. But order among distinct colors is assumed to not matter. This means I have to divide the total number of permutations by the number of ways I could permute balls of the same color. Then there would be a total of

$\dfrac{9!}{2!\cdot2!\cdot3!\cdot2!}=7,560$

ways of arranging the balls I had selected.