[proofplan] The event $\{S_n=k\}$ is the disjoint union of outcomes where exactly $k$ of the Bernoulli variables equal $1$. Each such pattern has probability $p^k(1-p)^{n-k}$ by independence, and there are $\binom nk$ patterns. [/proofplan] [step:Decompose by success sets] For $I\subset\{1,\ldots,n\}$ with $|I|=k$, let \begin{align*} E_I=\bigcap_{i\in I}\{X_i=1\}\cap\bigcap_{i\notin I}\{X_i=0\}. \end{align*} The events $E_I$ with $|I|=k$ are pairwise disjoint, and their union is $\{S_n=k\}$. [/step] [step:Compute each pattern] By independence and the Bernoulli law, \begin{align*} \mathbb P(E_I)=\prod_{i\in I}\mathbb P(X_i=1)\prod_{i\notin I}\mathbb P(X_i=0) =p^k(1-p)^{n-k}. \end{align*} There are $\binom nk$ subsets $I\subset\{1,\ldots,n\}$ of size $k$. Therefore \begin{align*} \mathbb P(S_n=k)=\sum_{|I|=k}\mathbb P(E_I)=\binom nk p^k(1-p)^{n-k}. \end{align*} [/step]