No. of ways to choose $n$ things from $n$ alike "a" things $n$ alike "b" things and $n$ different things. No. of ways to choose $n$ things from $n$ alike "a" things $n$ alike "b" things and $n$ different things.
Answer is $(n+2)2^{n-1}$. But how to prove it..? 
 A: Assume that we are choosing n objects from $\{a,\cdots,a,b,\cdots,b, c_{1},\cdots,c_{n}\}$
where there are n a's and n b's.
If we choose k elements from $\{a,\cdots,a,b,\cdots,b\}$, there are $k+1$ ways to do this;
and then we still have to choose $n-k$ elements from $\{c_1,\cdots,c_n\}$,
which can be done in $\binom{n}{n-k}=\binom{n}{k}$ ways.
Therefore the total number of choices is given by $\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}$.
Since $\displaystyle(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k$, $\;\;\;\;\;\displaystyle x(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^{k+1}$;  so 
differentiating gives 
 $(x+1)^{n-1}[(n+1)x+1]=\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}x^k$.
Now substituting $x=1$ gives that $\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}=2^{n-1}(n+2)$.

Another way to evalutate this would be to split up the sum as
$\displaystyle\sum_{k=0}^{n} k\binom{n}{k}+\sum_{k=0}^n \binom{n}{k}$
to get $n2^{n-1}+2^n=2^{n-1}(n+2)$.
A: At this stage the combinatorial problem is not clear. We give a suggestion for dealing with a sum quite a bit like the one you have at the end of the post. 
You seem to be asking for a closed form for something like
$$n\binom{n}{0}+(n-1)\binom{n}{1}+(n-2)\binom{n}{2}+\cdots +(1)\binom{n}{1}$$
Note that by the Binomial Theorem, 
$$(x+1)^n=\binom{n}{0}x^n+\binom{n}{1}x^{n-1}+ \binom{n}{2}x^{n-2}+\cdots+\binom{n}{n-1}x+\binom{n}{n}.$$
Differentiate. We get
$$n(x+1)^{n-1}=n\binom{n}{0}x^{n-1}+(n-1)\binom{n}{1}x^{n-2}+(n-2)\binom{n}{2}x^{n-3}+\cdots+ (1)\binom{n}{n-1}.$$
Put $x=1$. We get
$$n2^n =n\binom{n}{0}+(n-1)\binom{n}{1}+(n-2)\binom{n}{2}+\cdots+\binom{n}{n-1}.$$
