## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$10$
According to the binomial theorem, we have: $\displaystyle{n\choose k}=\dfrac{n!}{(n-k)! \ k!}$. Substitute $5$ for $n$ and $3$ for $k$ in the above formula. Therefore, $\dbinom{5}{3}=\dfrac{5!}{3! \ (5-3)!} \\ =\dfrac{5!}{3! \ 2!}\\=\dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(3\cdot 2 \cdot 1) (2 \cdot 1)}\\=10$