Q68 / 155 · WBJEE 2023 · Mathematics

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 the \(Q\) contains just one element more than \(P\) is

  • Correct answer
    A
    \({ }^{2 n} C_{n-1}\)
  • B
    \({ }^{2 n} C_n\)
  • C
    \({ }^{2 n} C_{n+2}\)
  • D
    \(2^{2 n+1}\)

Solution

Hint : Required number of ways \(\begin{aligned} & ={ }^{\mathrm{n}} \mathrm{C}_0 \cdot{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_1 \cdot{ }^{\mathrm{n}} \mathrm{C}_2+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \\ & ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}-1} \end{aligned}\)

Hence, the correct answer is option: 'A'.

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