We received many solutions to this problem,
but Charles from Clear Water Bay School in Hong Kong, Alex from
Longsands School and Maria from Wimbledon High Junior School took
the trouble to explain how they worked it out. Charles says:
The answer is $6$.If the green ball is in the first place,
there are two combinations i.e. red ball second, blue ball third or
blue ball second, red ball third.
That gives us two ways if the green ball
lands in the first place.
It is the same with the other two balls.
Therefore, $3 \times 2 = 6$ ways.
Here are the ways that
all three of them listed:
green, red, blue
green, blue, red
blue, red, green
blue, green, red
red, blue, green
red, green, blue