jmc

By the equation $$\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$$we have $$\left(\genfrac{}{}{0ex}{}{11}{4}\right)=\frac{11!}{4!7!}.$$$$\left(\genfrac{}{}{0ex}{}{11}{4}\right)=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(4\times 3\times 2\times 1)\times (7\times 6\times 5\times 4\times 3\times 2\times 1)}.$$This leaves us with just the first 4 terms of $11!$ in the numerator, and $4!$ in the denominator. Thus: $$\left(\genfrac{}{}{0ex}{}{11}{4}\right)=\frac{11!}{4!7!}=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}=\boxed{330}.$$