Let $C$ be $\left[n,k,2t+2\right]$ linear code above $GF(q)$.
Prove that there exists a coset of $C$ in $V(n,q)$ which contains two vectors with weight of $t+1$.
I didn't really know what to do with it.. I tried a lot and the only idea that seems worthwhile is to use pigeonhole principle to prove that there are more vectors with weight $t+1$ than cosets, than there must be a coset with two of these vectors, but it is really hard. to prove it, I have to prove that for $n \ge 2t+2$ :
$$ {n \choose t+1} > \sum_{i=1}^t {n \choose i}$$
But I don't how to prove it (and I don't even know if it's true). Thanks.
First thanks for Jyrki Lahtonen to help me solving. The solution is:
take a codeword $w$ with weight $2t+2$ (there must be such as the code is linear with minimal distance $2t+2$) and create from it a vector $y$ with weight $t+1$, that also $w(y+w)=t+1$. This is easy to do by taking $y$ to be all 0s where $w$ is 0s, all 0s in the first $t+1$ positions that are not 0 in $w$, and take the q-complement for each non-zero position in the last $t+1$ positions as such in $w$. This way the coset $y+C$ will contain $y$ itself (because the 0 vector must be a codeword and $y+0=y$) and $y+w$ (because we took $w$ as a codeword) which are diffrenet and both have weight of $t+1$. So the coset $y+C$ will be the necessary one.