Words of weight 5 in in Ternary Golay Code I'm not really good at doing this type of exercices. But I'd like to know how to prove that ther are 132 words of weight 5 in the Ternary Golay Code. I am not allowed to use the weight enumerator.
I tried to ask the same question in the global code $GF(3)^{11}$ but not succeeded. So I'm quite suck on this.
Any suggestion would be aweomse.

I assume that you are expected to answer this question using only the (big) piece of information that the ternary Golay code $G$ is a perfect code with covering radius $\rho=2$.
An attack (filling in the details as the OP solved the problem themself):


*

*The number of vectors of weight three in the space $GF(3)^{11}$ is equal to $\binom{11}3\cdot2^3=1320$. We can choose the three non-zero positions in $\binom{11}3$ ways, and each of those non-zero positions can have either $1$ or $2$ as the entry, so those three non-zero positions can be filled in $2^3$ different ways.

*The covering property implies that a vector $x$ of weight is within Hamming distance $\le 2$ of a unique word $w\in G$. The triangle inequality implies that the weight of $w$ must be in the interval $[1,5]$. The minimum distance of $G$ is five, so $w$ must be of weight five exactly.

*Given a codeword $w\in G$ of weight five, it is at distance two from exactly $\binom52=10$ vectors $x$ of weight three. This is because we get all such vectors $x$ by replaing two of the five non-zero components of $w$ with a zero. Note that the answer is independent of the choice of $w$.

*Let the number of codewords of weight five be $M$. In light of the previous bullet, between them they cover $10M$ vectors of weight three. Observe that there is no overlap, for if two distinct codewords were both within distance two of the same vector $x$, then the distance between them would be at most four in violation of the known minimum distance of $G$.

*Combining the first and the fourth bullet, we arrive at the equation $1320=10M$ implying $M=132$.

