Question about probability

45

10!/[(8!)(2!)]

Essentially, you have 10 choices for the first question you leave out, and 9 choices for the second question. This leads to 90 (10x9) combinations. But since it doesn't matter in which order you choose them, you've double counted (ie, you counted #1,#2 ans #2,#1 as different combinations), so divide by 2.

Njorl

 

Njorl's given you the correct answer, but to expand this is a simple cominatrics question, your calculator will probably have a function that allows you to work it out, combinations are given by:

[tex]_nC_r \equiv \left(\begin{array}{c}n\\r\end{array}\right) \equiv \frac{n!}{r!(n-r)!}[/tex]

that is n different things taken r at a time, so in this case n = 10 and r = 8.