$1365 = (10101010101)_2$
$(\text{total number of bits =} 11)$
Now if I keep $MSB = 0$, then from the remaining $10$ bits, I can select any $4$ bits ans set them as $1$. So total cases become $\binom{10}{4}$.
Now keep the $MSB = 1$, now we cannot replace $0$ on the second last bit from LHS with 1 as it will increase the number. This will apply to every $0$ as we move forward. So, now we have $10$ on the LHS and the third bit from LHS, we set to $0$, now we have $8$ bits remaning from which we have to select $3$ bits (Why $3$? as we are keeping MSB as $1$ and in total we only need $4$ bits to be set to $1$). Total cases will become $\binom{8}{3}$.
Similarly, we will move forward and will get $\binom{6}{2}$ and $\binom{4}{1}$.
So, the answer should be $\binom{10}{4} + \binom{8}{3} + \binom{6}{2} + \binom{4}{1}$
I think there is a typo in the $(C)$ option. They have the last term as $\binom{5}{1}$ instead of $\binom{4}{1}$.