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
 A: 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 $$2^n\binom {n+2-1}{1} - \binom n1 2^{n-1} \binom{n+1-1}{0} = (n+2)2^{n-1}$$
A: Your answer gives the  proposed result
$$
\eqalign{
  & \sum\limits_{j = 0}^n {\left( \matrix{
  j + 1 \cr 
  1 \cr}  \right)\left( \matrix{
  n \cr 
  n - j \cr}  \right)}  = \sum\limits_{j = 0}^n {\left( \matrix{
  j \cr 
  1 \cr}  \right)\left( \matrix{
  n \cr 
  n - j \cr}  \right)}  + \sum\limits_{j = 0}^n {\left( \matrix{
  j \cr 
  0 \cr}  \right)\left( \matrix{
  n \cr 
  n - j \cr}  \right)}  =   \cr 
  &  = \sum\limits_{j = 0}^n {\left( \matrix{
  j \cr 
  1 \cr}  \right)\left( \matrix{
  n \cr 
  j \cr}  \right)}  + \sum\limits_{j = 0}^n {\left( \matrix{
  n \cr 
  j \cr}  \right)}  = \sum\limits_{j = 0}^n {\left( \matrix{
  n \cr 
  1 \cr}  \right)\left( \matrix{
  n - 1 \cr 
  j - 1 \cr}  \right)}  + 2^{\,n}  =   \cr 
  &  = n\,2^{\,n - 1}  + 2^{\,n}  = \left( {n + 2} \right)2^{\,n - 1}  \cr} 
$$
where to obtain the last sum the "Trinomial Revision" has been applied
$$
\left( \matrix{
  n \cr 
  m \cr}  \right)\left( \matrix{
  m \cr 
  q \cr}  \right) = \left( \matrix{
  n \cr 
  q \cr}  \right)\left( \matrix{
  n - q \cr 
  m - q \cr}  \right)
$$
This results in $$ 1,3,8,20,48, \cdots$$
The above is the result when "combinations" means
number of sub-multisets of cardinality $n$ from the multiset $\{a \times n,\; b \times n,\; 1,\, 2,\, \cdots,n\}$
as explicitated by the z-Transform provided in previous answer.
That means that the order of the characters is not taken into account.
If instead "combinations" means
number of words of length $n$ from the alphabet $\{a \times n,\; b \times n,\; 1,\, 2,\, \cdots,n\}$
then the order of the characters is taken into account.
In this case, we have that $k$ numeric characters can be choosen 
in ${n^{\,\underline {\,k\,} } } =n(n-1)\cdots(n-k+1)$ ways.
In the remaining places we have a binary subword of length $n-k$.
Then the number of words will be
$$
\eqalign{
  & \sum\limits_{k = 0}^n {n^{\,\underline {\,k\,} } 2^{\,n - k} }  = \sum\limits_{k = 0}^n {{{n!} \over {\left( {n - k} \right)!}}2^{\,n - k} }  = n!\sum\limits_{k = 0}^n {{{2^{\,k} } \over {k!}}}  =   \cr 
  &  = e^{\,2} n!Q\left( {n,2} \right) \cr} 
$$
$Q$ being the Regularized Incomplete Gamma function.
This returns
$$
1,3,10,38,168, \cdots 
$$
