Given $n~1s$ and $n~0s$. The first digit of the number created by rearrangement of these $2n$ digits cannot be $0$. Find the sum. Let $n$ be a fixed positive integer. Find the sum of all positive integers with the following property: In base $2$, it has exactly $2n$ digits consisting of $n~1s$ and $n~0s$. The first digit cannot be 0.  
I know there are ${2n \choose n-1}$ such numbers. But, how can I find the sum. Please help. Thank you.
 A: There are $\binom{2n-1}{n-1}$ ways to rearrange the $n-1$ other ones and $n$ zeros.
Considering that the first digit is always $1$, each digit can be $1$ in exactly $\binom{2n-2}{n-2}$ ways (the number of permutations of the other $n-2$ ones and $n$ zeros) so that the sum would be
$$
\begin{align}
&\text{}\overbrace{\binom{2n-1}{n-1}}^{\text{number of integers}}\quad\overbrace{\vphantom{\binom{n}{n}}2^{2n-1}}^{\text{value of highest bit}}+\overbrace{\binom{2n-2}{n-2}}^{\begin{array}{c}\text{number of times}\\\text{each other bit is $1$}\end{array}}\quad\overbrace{\vphantom{\binom{n}{n}}\left(2^{2n-2}+2^{2n-3}+\dots+1\right)}^{\text{sum of the values of the other bits}}\\[9pt]
&=\frac{2n-1}{n-1}\binom{2n-2}{n-2}2^{2n-1}+\binom{2n-2}{n-2}\left(2^{2n-1}-1\right)\\[9pt]
&=\left(\frac{3n-2}{n-1}2^{2n-1}-1\right)\binom{2n-2}{n-2}
\end{align}
$$

Example
For $n=3$, we have $\left(\frac72\cdot2^5-1\right)\binom{4}{1}=444$ and
$$
\binom{5}{2}\text{ integers }
\left\{\begin{array}{}
100011=35\\
100101=37\\
100110=38\\
101001=41\\
101010=42\\
101100=44\\
110001=49\\
110010=50\\
110100=52\\
111000=56
\end{array}\right.\\
\qquad\qquad\qquad\text{total}=444
$$
