Prediction of index in Lexicographical ordering of binary sequences?

Question

Suppose all $n$ sequences of 0's and 1's were displayed in Lexicographical order. Can the index of sequences that sum to $k$ be predicted?

Answer 1

$\textbf{Better stated question}$: given all possible sequences of $n$ coin flips (a total of $2^n$ many), arrange them lexicographically. Can we find the index of those sequences that have $k$ heads?

$\textbf{Solution}:$ First let us consider a small case example where $n = 10$ and $k = 4$: \begin{cases} 0000001111 & = 2^{3} + 2^{2} + 2^1 + 2^0 \quad (\text{all $k$ 1's are at the end})\\ 0000010111 & = 2^{4} + 2^{2} + 2^1 + 2^0 \quad (k-1\text{ 1's are at the end; one 1 popped to the left})\\ 0000011011 & = 2^{4} + 2^{3} + 2^1 + 2^0 \quad (k-2\text{ 1's are at the end; one 1 popped to the left})\\ 0000011101 & = 2^{4} + 2^{3} + 2^2 + 2^0\\ 0000011110 & = 2^{4} + 2^{3} + 2^2 + 2^1\\ 0000100111 & = 2^{5} + 2^{2} + 2^1 + 2^1\\ 0000101011 & = 2^{5} + 2^{3} + 2^1 + 2^0\\ & \vdots\\ 1111100000 &= 2^{10} + 2^{9} + 2^8 + 2^7 \quad (\text{all} \ k \ \text{1's are to the left}) \end{cases}

Note for the equality I'm just translating binaries into decimals. For the general case, if $k > n$ then no such sequence exist, so assume $k \leq n$. Then the sequence we desire is \begin{cases} 0....01...1 & = 2^{k-1} + 2^{k-2} + 2^{k-3} +... 2^0 \quad (\text{all $k$ 1's are at the end})\\ 0...01011...1 & = 2^{k} + 2^{k-2} + 2^{k-3} + ... + 2^0 \quad (k-1\text{ 1's are at the end; one 1 popped to the left})\\ & \vdots\\ 1...10....0 &= 2^n + 2^{n-1} + ... + 2^{n-k+1} \end{cases}

so the answer is yes.