In a six-page homework assignment, we need to place four figures. The order of the figures is fixed, and a maximum of two figures can be on one page. In how many ways can this be done? (We do not consider the position of the figures on a page.)
We break down the task into three cases:

1. Each figure is on a separate page;
2. Two (consecutive) figures are on one page, the other two figures are on separate pages;
3. There are 2 figures on each of 2 pages.

In the first case, we choose 4 out of the 6 pages for the figures: this can be done in $\binom{6}{4}=\frac{6!}{4!\cdot 2!}=15$ ways.

In the second case, we choose 3 pages out of 6 for the figures, which can be done in $\binom{6}{3}$ ways. Then, from the 4 figures, we can place two on one page in 3 ways, which results in a total of $3 \cdot\binom{6}{3}=3 \cdot \frac{6!}{3!\cdot 3!}=60$ possibilities.

In the third case, we choose 2 pages out of 6, which is the same as the first case, or 15 possibilities. In total, there are $15+60+15=90$ ways to arrange the figures on the pages under the given conditions.