Identical Obects in Permutation and combination 
There are $2$ identical white balls , $3$ identical red balls and $4$ green balls of different shades. The number of ways in which the balls can be arranged in a row so that at least one ball is separated from the balls of the same colour. 

My Approach : We can find the answer using (Total Arrangements) $-$ (No ball is separated ) 
By this total number of arrangements are $$\frac{9!}{2!3!4!}$$ 
No ball is separated is $3!$ as the balls are identical.
But the answer is coming wrong.
Please tell the correct approach.
 A: There is a total of $\tfrac{9!}{2!3!}$ distinct ways to arrange the coloured balls, since the green balls are not identical.   (They are different shades of green !) $$\square\square\;\color{red}{\blacksquare\blacksquare\blacksquare}\;\color{#90EE90}\blacksquare\color{#71BC78}\blacksquare\color{lime}\blacksquare\color{#228B22}\blacksquare$$
To keep balls grouped so that none are separated from another of the same colour, we need to keep the identical whites stuck together as a pair, and the identical reds as a triplet, and the distinct greens must either form two separated pairs or a single group of four.   Count the ways:


*

*First arrange the reds to the left or right of the whites, leaving three spaces (before, between, and after) the blocks, where the greens may be placed

*Then either:


*

*Arrange all four greens into a single block and select one of the spaces to place them, or

*Take one specific shade of green, say $\color{#90EE90}\blacksquare$, and select one other green to arrange them as one pair, and arrange the two remaining greens as a second pair; then select one of three spaces for the first pair, and select one of the two remaining spaces for the second pair.


A: Possible patterns for no ball of any color being "single" are
$WWRRRGGGG$ with $3!$ permutations of blocks of colors
$\uparrow WW \uparrow RRR \uparrow$ with two blocks of $GG$ placed at the uparrows in $\binom32$ ways,
with $\frac{4!}{2!}$ permutations of blocks of colors,
thus unfavorable ways = $3! + \binom32\cdot\frac{4!}{2!} = 42$
Subtract this from the total ways

ADDED
If the interpretation is that the green balls are distinct, it is easy to modify the above as:
Total ways $=\dfrac{9!}{2!3!} = 4!\times \dfrac{9!}{2!3!4!}$
Unfavorable ways $= 4!3! + 4!\binom32\cdot2!$
A: Actually there are more than $3!$ ways that no balls can be separated, because the $4$ green balls could potentially be split into two disjoint pairs of green balls. Count those cases too.
