When you are told to compute all $k$-th partial derivatives of a function $f$ of $n$ variables $x_j$ then formally there are $n^k$ of them: For each word $(j_1,j_2,\ldots, j_k)\in [n]^k$ you differentiate first with respect to $x_{j_1}$, then with respect to $x_{j_2}$, $\ldots$, and finally with respect to $x_{j_k}$. On the left side of your equation you have the sum of all these $n^k$ derivatives (at the point ${\bf a}$), each multiplied with the corresponding monomial $x_{j_1}\ x_{j_2}\cdots\ x_{j_k}$.
Any two mixed partial derivatives where each variable has been affected the same number of times coincide, and at the same time the corresponding monomials $x_{j_1}\ x_{j_2}\cdots\ x_{j_k}$ coincide. Therefore among the $n^k$ terms on the left of your equation there are a lot of them equal. On the right side equal terms have been collected. The multiindex $\alpha=(\alpha_1,\alpha_2,\ldots, \alpha_n)$ of weight $|\alpha|=k$ encodes how often each of the individual variables $x_i$ is affected: For each $i\in[n]$ you should differentiate $\alpha_i$ times with respect to $x_i$ resp. take the factor $x_i^{\alpha_i}$ into the accompagning monomial. Now it is an elementary combinatorial fact that for each multiindex $\alpha$ of weight $k$ there are exactly $k!/\alpha!$ terms on the left side which correspond to this $\alpha$.
By the way: The large expressions considered here are generated when you develop the auxiliary function
$$\phi(t)\ :=\ f({\bf a} + t{\bf x})\qquad (t\in{\mathbb R})$$
into a Taylor series with respect to the real variable $t$. In order to compute $\phi^{(k)}(t)$ you have to apply the chain rule $k$ times in succession, and each time the formal number of terms is multiplied by $n$.