二项式反演

公式

公式 1

\[f(n) = \sum_{i = 0}^n (-1)^{n - i} \binom{n}{i} g(i) \Leftrightarrow g(n) = \sum_{i = 0}^n \binom{n}{i} f(i) \]

证明:

\[\because g(i) = \sum_{j = 0}^i (-1)^j \binom{i}{j} f(j) \\ \therefore f(n) = \sum_{i = 0}^n (-1)^{n - i} \binom{n}{i} \sum_{j = 0}^i \binom{i}{j} f(j) \]

将 \(\binom{n}{i}\) 乘进去:

\[f(n) = \sum_{i = 0}^n (-1)^{n - i} \sum_{j = 0}^i \binom{n}{i} \binom{i}{j} f(j) \]

容易发现 \(\binom{n}{i} \binom{i}{j} = \binom{n}{j} \binom{n - j}{i - j}\) :

\[f(n) = \sum_{i = 0}^n (-1)^{n - i} \sum_{j = 0}^i \binom{n}{j} \binom{n - j}{i - j} f(j) \\ = \sum_{j = 0}^n \sum_{i = j}^n (-1)^{n - i} \binom{n}{j} \binom{n - j}{i - j} f(j) \\ = \sum_{j = 0}^n \binom{n}{j} f(j) \sum_{i = j}^n (-1)^{n - i} \binom{n - j}{i - j} \]

令 \(k = i - j\) :

\[f(n) = \sum_{j = 0}^n \binom{n}{j} f(j) \sum_{k = 0}^{n - j} \binom{n - j}{k} (-1)^{n - j - k} 1^k \]

又由二项式定理, \((1 - 1)^x = \sum_{i = 0}^x \binom{x}{i} (-1)^i 1^{x - i}\) :

\[f(n) = \sum_{j = 0}^n \binom{n}{j} f(j) (1 - 1)^{n - j} = \binom{n}{n} f(n) = f(n) \]

\(\mathcal{Q.\ E.\ D.}\)

模型

"钦定"类问题

至少 \(\Leftrightarrow\) 恰好

\[f(x) = \sum_{i = x}^n \binom{i}{x} g(i) \Leftrightarrow g(x) = \sum_{i = x}^n (-1)^{i - x} \binom{i}{x} f(i) \]

其中 \(f(x)\) 为至少 \(x\) 个限制的方案数, \(g(x)\) 为恰好 \(x\) 个限制的方案数.

至多 \(\Leftrightarrow\) 恰好

\[f(x) = \sum_{i = 0}^x \binom{n - i}{n - x} g(i) \Leftrightarrow g(x) = \sum_{i = 0}^x (-1)^{x - i} \binom{n - i}{n - x} f(i) \]

在"钦定"意义下, "至多满足 \(x\) 个限制" \(\Leftrightarrow\) "至少不满足 \(n - x\) 个限制"