Chapter 11 - Section 11.5 - The Binomial Theorem - Exercise Set - Page 664: 52

$12$

Since $\left(\begin{array}{c} n\\ r\end{array} \right)=\dfrac{n!}{r!(n-r)!}$, then, $\left(\begin{array}{c} 12 \\ 11 \end{array} \right)$ is equal to \begin{array}{l}\require{cancel} \dfrac{12!}{11!(12-11)!} \\\\= \dfrac{12!}{11!(1)!} \\\\= 12 \end{array}