Permutation of different coloured balls There are 2 identical white balls, 3 identical red coloured balls and 4 green balls of different shades. Find the number of ways in which the balls can be arranged so that no two balls of the same colour appear together.
My attempt at the solution :
Total ways of arranging balls without restriction = $\frac{9!}{2!3!}$. 
Total ways in which balls can be arranged so that all same coloured balls are together = $6!$.
Therefore, the number of ways so that at least one ball of same colour is separated = $\frac{9!}{2!3!}$ - $6!$.
Where have I gone wrong?
 A: You are incorrect because it says "no two balls of the same colour appear together" whose contradiction is "at least two balls of the same color appear together." What you have calculated is "all balls of the same colour appear together" which is not the opposite of what they want. 
A: As @NL628 stated, the error you made is that you subtracted the number of ways in which all same coloured balls are together from the total number of permutations rather than the number of arrangements in which at least two balls of the same colour are together.
You correctly calculated the total number of arrangements.
To count arrangements in which at least two balls of the same colour are together, we can use the Inclusion-Exclusion Principle.
A pair of balls of the same colour are together:  This can occur in two ways.


*

*The two white balls are adjacent.

*Two of the red balls are adjacent.


Two white balls are adjacent:  We have eight objects to arrange: $G_1$, $G_2$, $G_3$, $G_4$, $R$, $R$, $R$, $WW$, where $G_k$ denotes the $k$th shade of green.  We choose three of the eight positions for the red balls.  The remaining five distinct objects can be placed in the remaining five positions in $5!$ orders.  Hence, the number of such arrangements is 

 $$\binom{8}{3}5! = \frac{8!}{3!5!} \cdot 5! = \frac{8!}{3!}$$

Two of the red balls are adjacent:  We again have eight objects to arrange:  $G_1$, $G_2$, $G_3$, $G_4$, $RR$, $R$, $W$, and $W$.  We choose two of the eight positions for the two white balls.  The remaining six distinct objects can be placed in the remaining six positions in $6!$ ways.  Hence, the number of such arrangements is 

 $$\binom{8}{2}6! = \frac{8!}{2!6!} \cdot 6! = \frac{8!}{2!}$$

Two pairs of balls of the same colour are together:  This can also occur in two ways.  


*

*Two disjoint pairs of adjacent balls of the same colour:  This means that there is a pair of adjacent red balls and a pair of adjacent white balls.

*Two overlapping pairs of adjacent balls of the same colour:  This means that there are three consecutive red balls.


Two disjoint pairs of adjacent balls of the same colour:  We have seven objects to arrange: $G_1$, $G_2$, $G_3$, $G_4$, $RR$, $R$, $WW$.  Each of the objects are distinct, so they can be arranged in 

 $7!$ 

ways.
Two overlapping pairs of adjacent balls of the same colour:  We again have seven objects to arrange:  $G_1$, $G_2$, $G_3$, $G_4$, $RRR$, $W$, $W$.  We select two of the seven positions for the white balls, then place the remaining five objects in the remaining five positions in $5!$ ways.  Hence, there are 

 $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$

such arrangements.
Three pairs of adjacent balls of the same colour:  This means the two white balls are adjacent and the three red balls are adjacent.  Hence, we have six objects to arrange: $G_1$, $G_2$, $G_3$, $G_4$, $RRR$, $WW$.  Since they are distinct, they can be arranged in 

 $6!$ 

ways.  
By the Inclusion-Exclusion Principle, the number of arrangements in which there are at least two balls are of the same colour are adjacent is 

 $$\frac{8!}{3!} + \frac{8!}{2!} - 7! - \frac{7!}{2!} + 6!$$

If you subtract this quantity from the total number of arrangements, you will find the number of arrangements in which no two balls of the same colour are together.
