Let the balls be Red Blue and Yellow. My method requires that the ball be thought of as different
the pairs are RB, RY and BY so there are 3 ways to pair them.
Number of ways the balls can go into the cups is 10*10*10 = 1000
Number of ways to get Blue and Red together and Yellow somewhere else = 10*1*9=90
there are 3 possible pairings so that is 90*3=270
So P(exactly 2 in the same cup) = 270/1000 = 27/100
Another way to look at it is by subtraction:
the number of ways they can all be in different cups is 10*9*8 = 720
The number of ways they can all be in the same cup is 10*1*1 =10
so the number of ways that 2 will be in a cup is 1000-720-10 = 270
the answer is the same so that is conclusive enough for me.