(b) Suppose we have \(n\) people and we want to make a committee of arbitary size. The amount of ways we can choose a committee of 0 people is \(\dbinom{n}{0}\). The amount of ways we can choose a committee of 1 person is \(\dbinom{n}{1}\). In general, the amount of ways we can choose a committee of k people is \(\dbinom{n}{k}\). Then, the total would be
\(\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + \cdots + \dbinom{n}{n}\)
ways. However, notice that this is also equivalent to choosing whether or not a person should be on the committee. Each person has 2 choices - be on the committee or don't. Then, there are \(2^n\) total ways. Conclusively, \(\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + \cdots + \dbinom{n}{n}\)=\(2^n\).