Combination of $n$ letter out of $3n$. 
Show that the number of combinations of $n$ letters out of $3n$ letters of which $n$ are $a$s , $n$ are $b$s and the rest are unequal is $(n+2)2^{n-1}$.

My approach...
The number of ways of distributing $n$ distict object in $k$ distinct box is $\binom{n+k-1}{k-1}$ now when $k=2$ the logic stands as $\binom{n+2-1}{2-1}$=$\binom{n+1}{1}$.
Selection of $j$ distinct letter out of $n$ is $\binom{n}{j}$.
Hence my answer will be $\sum_{j=0}^{n} \binom{j+1}{1}\binom{n}{n-j}$.
Please help me with the answer

Hint: The required number is coefficient of $x^n$ in $$[(1+x+x^2+\cdots x^n)^2(1+x)^n].$$
Do some simplification and find that the coefficient of $x^n$ is $$2^n\binom {n+2-1}{1} - \binom n1 2^{n-1} \binom{n+1-1}{0} = (n+2)2^{n-1}$$
