## Precalculus (6th Edition) Blitzer

The order in which collections of 4 books are selected does not matter. Thus, this is a problem of selecting 4 books from a group of 12 books. We need to find the number of combinations of 12 things taken 4 at a time. We use the formula, ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Where $n=12,r=4$ \begin{align} & {}_{12}{{C}_{4}}=\frac{12!}{\left( 12-4 \right)!4!} \\ & =\frac{12!}{8!\times 4!} \end{align} Solve further, \begin{align} & {}_{12}{{C}_{4}}=\frac{12\times 11\times 10\times 9\times 8!}{4\times 3\times 2\times 1\times 8!} \\ & =11\times 5\times 9 \\ & =495 \end{align} Thus, there are 495 ways to take different collections of 4 books from 12 books.