I have a problem involving characters of a certain $\mathbb C$$G$-module, where $G = S_n$. The module is the vector-space $V$ with basis $\{v(I) | I\subset\{1,...,n\}，|I| = k\}$, where $k\le n$. With $\chi$ the character induced by the natural action of $S_n$ on $V$, the question asks to prove,
$(\chi,\chi )=min(k,n-k)+1$
But I am struggling to find a way to tackle this. My attempts have involved direct calculation but I didn't find a nice way to characterise the sum. I have also tried to deduce through applying similar ideas to those that are use when dealing with the normal permutation character. This included that the inner product is related to fixed points, or even related to the orbits of $V\times V$. But in all cases I can't see any way to characterise the orbits or fixed points.
I believe this problem has some connections with quite deep representation theory, but I would like a more combinatorial approach to the problem, seeing as I have only had a basic introduction to representation and character theory!
Without loss of generality $k\leq n/2$ (otherwise replace $k$ by $n-k$, the modules are isomorphic via the complement map). As Jyrki points out, it is enough to show that there are exactly $k+1$ orbits on $V\times V$. Certainly if $(a,b)$ and $(c,d)$ lie in the same orbit then $|a \cap b |=|c \cap d|$. This intersection can have any size between $0$ and $k$, because $k\leq n/2$. So there are at least $k+1$ orbits, and if we show all pairs $(a,b)$ with fixed $|a \cap b|$ lie in the same orbit then there are exactly $k+1$.
Suppose $|a \cap b| = r \leq k$. The subsets $a \cap b, a\setminus (a \cap b), b \setminus (a \cap b)$ are disjoint so we can choose a permutation $\pi \in S_n$ such that $\pi$ maps $a \cap b$ to $\{1,\ldots,r\}$, $a \setminus (a\cap b)$ to $\{r+1,\ldots,k\}$ and $b \setminus (a\cap b)$ to $\{ k+1,\ldots,k+(k-r)\}$. Then $$ \pi (a,b) = (\{1,\ldots,k\}, \{1,\ldots,r\}\cup \{k+1\ldots,2k-r\})$$ and every such pair $(a,b)$ therefore lies in the orbit of $(\{1,\ldots,k\}, \{1,\ldots,r\}\cup \{k+1\ldots,2k-r\})$.