A is a set containing n elements. A subset P₁ of A is chosen. The set A is reconstructed by replacing the elements of P₁. Next, a subset P₂ of A is chosen and again the set is reconstructed by replacing the elements of P₂. In this way, m (>1) subsets P₁ , P₂, ..., P_m of A are chosen. The number of ways of choosing P₁, P₂, ..., P_m is

Let $\mathrm{A}=\left\{{\mathrm{a}}_1,{\mathrm{a}}_2,\dots ,{\mathrm{a}}_{\mathrm{n}}\right\}$. For each ${\mathrm{a}}_{\mathrm{i}}(1\le \mathrm{i}\le \mathrm{n})$, we have either ${\mathrm{a}}_{\mathrm{i}}\in {\mathrm{P}}_{\mathrm{j}}\text{ or }{\mathrm{a}}_{\mathrm{i}}\notin {\mathrm{P}}_{\mathrm{j}}(1\le \mathrm{j}\le \mathrm{m})$. That is, there are 2^m choices in which ${\mathrm{a}}_{\mathrm{i}}(1\le \mathrm{i}\le \mathrm{n})$ may belong to the P_j's. One of these, there is only one choice, in which ${\mathrm{a}}_{\mathrm{i}}\in {\mathrm{P}}_{\mathrm{j}}$ for all j = 1, 2, ..., m which is not favourable for ${\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \cdots \cap {\mathrm{P}}_{\mathrm{m}}\text{ to be }\mathrm{\varphi }$. Thus, ${\mathrm{a}}_{\mathrm{i}}\notin {\mathrm{P}}_1\cap {\mathrm{P}}_2\cap \cdots \cap {\mathrm{P}}_{\mathrm{m}}\text{ in }{2}^{\mathrm{m}}-1$ ways.

Since there are n elements in set A, the total number of choices is ${\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}$

Also, there is exactly one choice, in which, ${\mathrm{a}}_{\mathrm{i}}\notin {\mathrm{P}}_{\mathrm{j}}$ for all j = 1, 2, …, m which is not favourable for ${\mathrm{P}}_1\cup {\mathrm{P}}_2\cup \cdots \cup {\mathrm{P}}_{\mathrm{m}}$ to be equal to A.

Thus, a_i can belong to ${\mathrm{P}}_1\cup {\mathrm{P}}_2\cup \cdots \cup {\mathrm{P}}_{\mathrm{m}}\text{ in }\left(2^{\mathrm{m}}-1\right)$ ways.

Since there are n elements in set A, the number of ways in which ${\mathrm{P}}_1\cup {\mathrm{P}}_2\cup \cdots \cup {\mathrm{P}}_{\mathrm{m}}$ an be equal to A is ${\left(2^{\mathrm{m}}-1\right)}^{\mathrm{n}}$.