caesarkim
there are 10 questions. i can choose only exact 8 questions.

what is the total combination?

Njorl
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

jcsd
$$_nC_r \equiv \left(\begin{array}{c}n\\r\end{array}\right) \equiv \frac{n!}{r!(n-r)!}$$