Equation for generating integers for n-bit binary strings with k bits set to 1 Assume there is an $n$-bit binary string where $k$ bits of it are set to $1$. We can show that it results in $(^n_k)$ number of binary strings with k bits set to $1$. 
How to define an equation to generate these $(^n_k)$ numbers (in integers).
For example, consider the situation of $n=3$ and $k=2$. Then we can generate the following binary sequences,
0 1 1 = 3
1 0 1 = 5
1 1 0 = 6

So, how can a function, $F(n,k)$ be defined so that, $F(3,2)$ generates, $3,5,$ and $6$ as the answers?
 A: You probably want a function $F(n,k,i)$ which gives the $i^{th}$ number in increasing order that has $n$ binary bits of which $k$ are $1$.  As you say, there are $n \choose k$ of them and it will be convenient to let $i$ range from $0$ to ${n \choose k}-1$.  There are ${n-1 \choose k}$ that start with a $0$ in the most significant bit and ${n-1 \choose k-1}$ that start with a $1$, so we have a simple recursion
$$F(n,k,i)=\begin {cases} 0&k=0\\
2^k-1&n=k\\F(n-1,k,i)&i \lt {n-1 \choose k}\\ 2^{n-1}+F\left(n-1,k-1,i-{n-1 \choose k}\right)& i \ge {n-1 \choose k} \end {cases}$$
because if $i$ is small, we write a $0$ for the first bit and want the $i^{th}$ number that has $n-1$ bits of which $k$ are $1$.  If $i$ is large, we write a $1$, which contributes $2^{n-1}$ to the value and subtract from $i$ the number of numbers that have a $0$ in the first bit.  
Added:  for $n=5,k=3$ you start with ${4 \choose 3}=4$ words that start with $0$ and follow with ${4 \choose 2}=6$ that start with $1$.  That gives
$$00111\\01011\\01101\\01110\\10011\\10101\\10110\\11001\\11010\\11100$$
