| Think it through - the student can pick any one of eight problems for his first, then after he completes that he can choose any one of the 7 remaining for the next etc. So total number of ways that 5 problems can be chosen out of 8 is 8*7*... *4. Now in this case the order of choice doesn't matter - for example, selecting problem number 3 first and number 4 second is the same as selecting 4 first then 3 second. Hence you divide the above by the number of ways that the 5 problems he selects could be arranged, which is 5*4*...*1. The result is known as C(8,5), the combination of 8 things chosen 5 at a time. | 
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