Question 1096554
<br>You need to choose either 1, 2, 3, 4, or 5 of the 10 problems.  The numbers of ways to do that are "10 choose 1", "10 choose 2", ..., and "10 choose 5".<br>
The easiest way to find those numbers is to look at the 5th row of Pascal's Triangle, which begins
1, 10, 45, 120, 210, 252, ...<br>
Those numbers are 10 choose 0, 10 choose 1, ..., 10 choose 4, and 10 choose 5.<br>
So the number of ways to choose at least 1 and at most 5 problems is
10+45+120+210+252 = 637