In problems like this I find it a little easier to break it down into multiple cases:
- All the balls are different colours.
- One pair of chosen balls are the same colour.
- Two pairs of chosen balls are the same colour.
- Three pairs of chosen balls are the same colour.
How many combinations are there in case 1? Well, just one. There are six colours, and balls of the same colour are considered
How many combinations are there in case 2? Say that $C_1$ is the colour ball which is paired. Then there are 4 balls, all of different colours, remaining. That means there's one colour missing. There are 5 ways for that to happen, because there are 5 colours left when $C_1$ is the colour of the ball which is paired. This should be a clue as to how many total combinations there are for case two, given that I've only considered here the possibility of $C_1$ being the colour of the paired ball.
In case three, we use similar logic. Say that $C_1$ and $C_2$ are the colours which are paired. There are two balls of different colours remaining, and two colours missing. There's only a limited number of ways to have two colours remaining. You should be able to do that. Remember again that this is only the case where $C_1$ and $C_2$ are the colours of the paired balls.
The last case is where three pairs are the same colour. There are a number of ways which that can happen, and it's the same as the number of ways of picking 3 objects from 6.