B5. Raymond has five coins. On the head side of each coin is the number 1. On the tail side are the fractions $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$, and $\frac{1}{6}$, respectively. Since each coin can either land heads up or tails up, there are 32 ways to place the five coins on the table. For each of these 32 ways, Raymond multiplies the five numbers that are facing up and writes down the result.

When Raymond finally adds up these 32 numbers, what is the result?

## C-problems

For the C-problems, the solution method is as important as the answer; you must clearly describe how you solved it.

Work out each C-problem on a separate sheet of paper. Each correctly worked out C-problem is worth 10 points.

Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper (for each problem!) as well.
B5. The outcomes that Raymond can obtain correspond exactly to the terms in the expansion of the product $\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{6}\right)$.

The total sum is then naturally

$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{6}\right)=\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}=\frac{7}{2} .
$$

## C-problems