Find how many committees with a chairman can be chosen from a set of n persons. Hence or otherwise prove that
$${n \choose 1} + 2{n \choose 2} + 3{n \choose 3} + ...... + n{n \choose n} = n2^{n-1}$$
1. **Counting Committees with a Chairman:**
   - To form a committee with a chairman from a set of \( n \) persons, we can choose any subset of the \( n \) persons (including the empty set) and then select one person from this subset to be the chairman.
   - The number of ways to choose a subset of \( k \) persons from \( n \) persons is given by \( \binom{n}{k} \).
   - For each subset of \( k \) persons, there are \( k \) ways to choose the chairman.
   - Therefore, the number of ways to form a committee of size \( k \) with a chairman is \( k \binom{n}{k} \).
   - Summing over all possible sizes of the committee (from 1 to \( n \)), the total number of ways to form a committee with a chairman is:
     \[
     \sum_{k=1}^{n} k \binom{n}{k}
     \]

2. **Proving the Identity:**
   - We start with the binomial theorem:
     \[
     (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
     \]
   - Differentiate both sides with respect to \( x \):
     \[
     \frac{d}{dx} (1 + x)^n = \frac{d}{dx} \left( \sum_{k=0}^{n} \binom{n}{k} x^k \right)
     \]
   - The left-hand side becomes:
     \[
     n(1 + x)^{n-1}
     \]
   - The right-hand side becomes:
     \[
     \sum_{k=1}^{n} k \binom{n}{k} x^{k-1}
     \]
   - Therefore, we have:
     \[
     n(1 + x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} x^{k-1}
     \]
   - Setting \( x = 1 \) in the above equation:
     \[
     n(1 + 1)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} 1^{k-1}
     \]
   - Simplifying, we get:
     \[
     n \cdot 2^{n-1} = \sum_{k=1}^{n} k \binom{n}{k}
     \]
   - Hence, we have proved that:
     \[
     \sum_{k=1}^{n} k \binom{n}{k} = n \cdot 2^{n-1}
     \]

\(\blacksquare\)

The final answer is \( \boxed{ n \cdot 2^{n-1} } \)