How many ways are there to make a binary string with more $1$'s than $0$'s such that the first two bits are $1$'s? How many ways are there to make a binary string of lenght $n$ with more $1$'s than $0$'s such that the first two bits are $1$'s ? (Call this number "$c$").
Let $c_0$ be the number of strings with more $0$'s, $c_1$ the number of strings with more $1$'s and $c_2$ the number of strings with the same number of both. For every $n$, $c_0+c_1+c_2=2^n$ and $c_0=c_1$ so $$c_1=\frac{2^n-c_2}{2}$$ How can $c_2$ be calculated? For odd $n$, $c_2=0$ but for even $n$ is it $\binom{2n}{n}$ ? I'm not really sure the "stars and bars" approach is well used here
Now how can the desired number be calculated (supposing everything above was right) ? If the string starts with $11$ then there are only $2^{n-2}$ options left so is $c=c_2/2^{n-2}$ ?
Please explain as claerly as possible how to get the correct solution (I'm very bad in combinatorics). Thanks
 A: Your thinking is along the right lines, but since the first two bits have to be a $1$, it is better to consider all strings of length $m=n-2$ where the number of $1$'s is at least the number of $0$'s minus $1$.
So, where $m$ is odd, so say $m = 2k+1$, you can have the number of $1$'s be exactly the number of $0$'s minus $1$, meaning that you have $k$ $1$'s and $k+1$ $0$'s, and there are ${m}\choose{k}$ such strings.  And since in half of all the $2^m$ strings you have more $1$ than $0$'s, that gives a total of 
$${{m}\choose{k}} + 2^{m-1}$$ 
strings of length $m$ with more $1$'s than $0$'s. 
Thus, as a function of the original $n$, if $n$ is odd, then there are:
$${{n-2}\choose{\frac{n-3}{2}}} + 2^{n-3}$$ 
strings of length $n$ that start with $11$ and that have more $1$'s than $0$'s.
When $m$ is even, say $m = 2k$, the number of $1$'s has to equal or exceed the number of $0$.  There are $m \choose k$ strings with an equal number of $1$'s and $0$'s, and in half of the rest you have more $1$'s than $0$'s, so that is $\frac{2^m-{m \choose k}}{2}$ such strings, for a total of 
$${m \choose k} + \frac{2^m-{m \choose k}}{2}$$
strings of length $m$ with more $1$'s than $0$'s. 
In terms of the original $n$, you thus have that if $n$ is even, then there are:
$${{n-2} \choose \frac{n-2}{2}} + \frac{2^{n-2}-{n-2 \choose \frac{n-2}{2}}}{2}$$
strings of length $n$ that start with $11$ and that have more $1$'s than $0$'s.
