## Precalculus (6th Edition) Blitzer

The order in which the four-person committee is selected does not matter. Thus, this is a problem of selecting 4 people from a group of 11 people. We need to find the number of combinations of 11 things taken 4 at a time. Apply the formula, ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Where $n=11,r=4$ \begin{align} & {}_{11}{{C}_{4}}=\frac{11!}{\left( 11-4 \right)!4!} \\ & =\frac{11!}{7!\times 4!} \end{align} Solving further, \begin{align} & {}_{11}{{C}_{4}}=\frac{11\times 10\times 9\times 8\times 7!}{4\times 3\times 2\times 1\times 7!} \\ & =11\times 10\times 3 \\ & =330 \end{align} Hence, there are 330 ways to select the four-person committee from a group of 11 people.