**Concept: The question tests on the application of counting**

**Solution:**

We have 2 White identical balls (WW), 3 Red identical balls (RR) and 4 different green balls (ABCD)

The total balls we have = 2 + 3 + 4 = 9

We want to find the number of ways when at least 1 ball is separated from balls of the same color

This is equal to The total arrangements possible - Total ways when balls of the same color are together.

Total Arrangements = 9!/2!∗3!=9∗8∗7!/12=6∗7!

Arrangements where all balls of the same colour are together: Let us consider 2 white as 1 unit, the 3 red as 1 unit and the 4 Green as the 3rd unit. Then the external arrangement = 3!

For the internal arrangement, we to consider the 4 green balls only = 4!

Arrangements, where similar colored balls are together = 3! * 4! = 6 * 4!

The required number of arrangement = 6 * 7! - 6 * 4! = 6(7! - 4!)**Option A**