From 4 red, 2 green and 3 white balls, how many selections consisting of 7 balls are possible, if each colour must be represented at least twice?

Question

From 4 red, 2 green and 3 white balls, how many selections consisting of 7 balls are possible, if each colour must be represented at least twice? Assume balls of the same colour are identical.

I thought I should do this in the following way:

$$^4C_2 \times \, ^2C_2 \times \, ^3C_2 = 18$$

But these are only six balls so how should I do it for seven?

$18 \times \, ^3C_1$? Choosing 1 from the remaining 3?

Also, what does it mean exactly that the balls are identical?

Answer 1

"Each colour must be represented at least twice". So take two balls of each colour. How many additional balls must you choose? And how many colours are there among the remaining balls?

Answer 2

The greens with you have finished if you need at least 2 balls of each colour. Now you need just one ball from 2 red and 1 white balls. Now the remaining solution is easy with intuition. But if you want a pure mathematical method then you can try star and bars method.
Using star and bars method we need to find non negative integral solutions of the equation $$a+b=1$$

Which is simply $\binom {2}{1}=2$