Is one-time-pad still secure if the number of 1's in the key is revealed to the attacker? For example, if $m = 10011$, $k = 11001$, $n=3$ (which is the number of 1's in k), $c = m \oplus k = 01010$. If $c$ and $n$ are revealed to the attacker, is this scheme still secure?
The security notion one usually considers for OTP is perfect secrecy, which informally means that the ciphertext does not reveal any information about the original message, regardless of the computational power of the adversary. It is already known that this requires that the key size must be equal to the plaintext size and that all keys are equiprobable. That way, for each ciphertext you could find a key that decrypts the ciphertext to any possible message. 

However, your proposal violates these principles. Assuming the key size (and plaintext size) is $S$, then the number of possible keys is not $2^S$ (i.e., all the possible keys of length $S$, as in the regular OTP), but $\binom{S}{n}$, which is much less than $2^S$. This has the consequence that for a given message $m$, not all possible ciphertexts are equiprobable, or conversely, for a given ciphertext not all possible messages are equiprobable (in fact, some of them have no probability).

Let's see it with your example. With regular OTP, for a length of 5 bits there are $2^5 = 32$ possible keys (as well as messages and ciphertexts). If an adversary gets $c$, there are 32 possible messages that correspond to that ciphertext. That is, the ciphertext is completely useless to the adversary. Now, in your proposal, for $n = 3$ there are $\binom{5}{3} = 10$ possible messages for a given ciphertext, since there are only 10 possible keys. The ciphertext $c$ and knowledge of $n$ makes possible to the adversary to deduce which 10 messages (out of the 32 possible) are related to the ciphertext. Therefore, he is gaining some knowledge about the original message.