# Counting arrangements in which at least one ball is separated from the balls of same colour

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

My Approach

First, I found the total arrangements of the balls in a row which came out to be:

$$\frac {9!}{2!3!}$$

Now, I took the case where all the balls are together.

So the total number of balls in this case is $= 3$

So total arrangements came out to be= $3! \times 4!$

So total arrangements is subtracting the balls together case from total arrangement case.

But somehow this answer is wrong even though it looks right to me....Can you tell me my mistakes here?

I presume the green balls count as different colors. To compute the number of arrangements with both the white balls together and all the red balls together, you can just group the whites into one ball and the reds into another. The number of arrangements is then $6!$, so the final answer is $$\frac {9!}{2!3!}-6!=29520$$