## Thinking Mathematically (6th Edition)

$450$ ways
The examine has a sequence of selections: 1. ... 8 out of a group of 10 multiple choice questions ... in ${}_{10}C_{8}$ ways 2. ... 3 out of a group of 5 open-ended problems... in ${}_{5}C_{3}$ ways By the Fundamental Counting Principle, Total ways= ${}_{10}C_{8}\cdot {}_{5}C_{3}$ ${}_{10}C_{8}=\displaystyle \frac{10!}{(10-8)!8!}=\frac{10\times 9}{1\times 2}=45$ ${}_{5}C_{3}=\displaystyle \frac{5!}{(5-3)!3!}=\frac{5\times 4}{1\times 2}=10$ Total ways= ${}_{10}C_{8}\cdot {}_{5}C_{3} =45\times 10=450$