Compute $\displaystyle \binom{11}{4}$.
By the equation $$ \dbinom{n}{r} = \frac{n!}{r!(n-r)!} $$we have $$ \binom{11}{4} = \frac{11!}{4!7!}. $$$$ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times1}{(4 \times 3 \times 2 \times 1)\times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times1)}. $$This leaves us with just the first 4 terms of $11!$ in the numerator, and $4!$ in the denominator.  Thus: $$ \binom{11}{4} = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \boxed{330}. $$