Let $j \in \mathbb{R}^n$ be a multi-index with $|j|=3$. Fix an index $l$, then there are how many $X^j$ with $j_l=0$?, more precise: I want to know what $ \text{#} \{ X^j : |j|=3, j_l \neq 0\}$ is.
So did the following: If $j$, $|j|=3$, then $X^j=X_{i_1} X_{i_2} X_{i_3}$ (for example $j_i=3$, then choose $i_1=i_2=i_3=i$), and $j_l=1$ means that $i_3=l$ already fixed, $X^j=X_{i_1} X_{i_2} X_l$. So if we disjunctly seperate it in the cases $j_l=1,2$ and $3$, we would have
- $\text{#} \{ X^j : |j|=3, j_l =1 \}= \text{#}\{ X_{i_1} X_{i_2} X_l : i_1,i_2 \neq l \}=(n-1)^2$ (place 2 stones $i_1, i_2$ on a line with $n-1$ free places (place l is "occupied") )
- $\text{#} \{ X^j : |j|=3, j_l =2 \}= \text{#}\{ X_{i_1} X_l^2 : i_1 \neq l \}=n-1$
- $\text{#} \{ X^j : |j|=3, j_l =3 \}= \text{#}\{ X_l^3 \}=1$
So in sum $n^2-n-1$. The problem is that, if I check this, it should equal $\text{#} \{X^j : |j|=3\} - \text{#} \{ X^j : |j|=3, j_l=0 \} = n^3 - \text{#} \{ X_{i_1} X_{i_2} X_{i_3} : i_1, i_2, i_3 \neq l \}=$ $=n^3-(n-1)^3$
But $n^3-(n-1)^3=1-3n+3n^2$, so I did something wrong somewhere.. Does somebody see the mistake(s)? (Or has another way to assure the result)