Theorem 9.34 Rudin

I don't understand what do we mean in $j_q$ and $j_p$. I also don't understand part $(b)$ of 9.34 theorem. I could n't understand this sentence: "det is a linear function of each of the column vectors $x_j$, if the others are held fixed" and there by from Where did $n!$ come?

For your first question: In the product of equation (82), we are iterating over all pairs of indices $(p,q) \in \{1, \dotsc, n\}$ such that $p < q$. Therefore $j_p$ and $j_q$ are a pair of elements from the set $\{j_1, \dotsc, j_n\}$ such that $p < q$, in other words, $j_p$ precedes $j_q$ in the ordered $n$-uple $(j_1, \dotsc, j_n)$. For example we take the couples $(j_1, j_2)$ and $(j_2, j_4)$, but not the couples $(j_2, j_1)$, $(j_1, j_1)$ or $(j_4, j_2)$.

For the second part: $\det$ is a linear map of each of the column vectors $x_j$ means that $$\det \begin{pmatrix} x_1 & \dots & (\lambda x_j + y_j) & \dots & x_n \end{pmatrix} = \\ \lambda \det\begin{pmatrix} x_1 & \dots & x_j & \dots & x_n \end{pmatrix} + \det\begin{pmatrix} x_1 & \dots & y_j & \dots & x_n \end{pmatrix} $$ for every scalar $\lambda \in \mathbb{R}$, for every column vectors $x_j, y_j \in \mathbb{R}^n$, and for any index $j \in \{1, \dotsc, n\}$.

The $n!$ comes from the fact that we are iterating over all the $n$-tuples $(j_1, \dotsc, j_n)$ with $1 \leq j_r \leq n$, for every $r \in \{1, \dotsc, n\}$, i.e. over all the permutations of the $n$-elements set $\{j_1, \dotsc, j_n\}$. A basic combinatorial argument gives $n!$ as the number of permutations of such set.