1.2.2 problem 18

Consider the right hand side of the equation. Since a committe chair can only be selected from the first group, there are $n$ ways to choose them. Then, for each choice of a committee chair, there are $\left(\frac{2n-1}{n-1}\right)$ ways to choose the remaining members. Hence, the total number of committees is $n\left(\frac{2n-1}{n-1}\right)$.

Now consider the left side of the equation. Suppose we pick $k$ people from the first group and $n-k$ people from the second group, then there are $k$ ways to assign a chair from the members of the first group we have picked. $k$ can range from $1$ to $n$ giving us a total of ${\sum }_{k=1}^{n}k\left(\frac{n}{k}\right)\left(\frac{n}{n-k}\right)={\sum }_{k=1}^{n}k{\left(\frac{n}{k}\right)}^2$ possible committees.

Since, both sides of the equation count the same thing, they are equal.