Ans: (A)
Hint: Ways of selections are
\(\mathrm{n}\) identical and no different \(=1\) way
\(n-1\) identical and one from different elements \(=1 \times n_{c_{1}}\)
0 identical rest from different \(=1 \times{ }^{n} \mathrm{C}_{n}\)
\(\sum=^{n} c_{0}+^{n} c_{1}+^{n} c_{2}+\ldots .+^{n} c_{n}=2^{n}\)