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(A) Let $|P| = k$. Then $Q$ must have $k+1$ elements.For a fixed $k$, the number of ways to choose $P$ is $^{n}C_k$.The number of ways to choose $Q$ i

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$^{2n}C_{n-1}$

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(A) Let $|P| = k$. Then $Q$ must have $k+1$ elements.For a fixed $k$, the number of ways to choose $P$ is $^{n}C_k$.The number of ways to choose $Q$ is $^{n}C_{k+1}$.Since $P$ and $Q$ are chosen independently, the total number of ways is the sum over all possible values of $k$ from $0$ to $n-1$:$\sum_{k=0}^{n-1} (^{n}C_k \cdot ^{n}C_{k+1})$Using the identity $^{n}C_k = ^{n}C_{n-k}$, we have:$\sum_{k=0}^{n-1} (^{n}C_{n-k} \cdot ^{n}C_{k+1})$This is the coefficient of $x^{n+1}$ in the expansion of $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$, which is $^{2n}C_{n+1}$.Note that $^{2n}C_{n+1} = ^{2n}C_{2n-(n+1)} = ^{2n}C_{n-1}$.

Let $A$ be a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen, and the set $A$ is reconstructed by replacing the elements of $P$. $A$ subset $Q$ of $A$ is chosen again. The number of ways of choosing $P$ and $Q$ such that $Q$ contains just one element more than $P$ is

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