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Hint: Arithmetic mean is the sum of a collection of numbers divided by the count of numbers in the collection. This definition will be useful in solving the problem.

\[mean = \dfrac{{0 \times 1 + 1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{1 + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}\]

We can write \[1\]as\[C\left( {n,0} \right)\], so that we can get a pattern and solve it easily.

\[ = \dfrac{{1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{C\left( {n,0} \right) + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}\]

\[ = \dfrac{{\sum\limits_{r = 1}^n {r \times C\left( {n,r} \right)} }}{{{2^n}}}\]

Answer \[ = \dfrac{{n \times {2^{n - 1}}}}{{{2^n}}} = \dfrac{n}{2}\]

Note: Make sure that you take the limits correctly otherwise you are likely to make mistakes in reaching the answer.

\[mean = \dfrac{{0 \times 1 + 1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{1 + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}\]

We can write \[1\]as\[C\left( {n,0} \right)\], so that we can get a pattern and solve it easily.

\[ = \dfrac{{1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{C\left( {n,0} \right) + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}\]

\[ = \dfrac{{\sum\limits_{r = 1}^n {r \times C\left( {n,r} \right)} }}{{{2^n}}}\]

Answer \[ = \dfrac{{n \times {2^{n - 1}}}}{{{2^n}}} = \dfrac{n}{2}\]

Note: Make sure that you take the limits correctly otherwise you are likely to make mistakes in reaching the answer.

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