Chapter 11 - Probability and Statistics - 11-8 Samples and Surveys - Practice and Problem-Solving Exercises - Page 730: 36

$_3C_3=1$

Using $ _nC_r=\dfrac{n!}{r!\text{ }(n-r)!} $ or the Combination of $n$ taken $r,$ then \begin{align*}\require{cancel} _3C_3&= \dfrac{3!}{3!\text{ }(3-3)!} \\\\&= \dfrac{3!}{3!\text{ }0!} \\\\&= \dfrac{3!}{3!\text{ }1} &\text{(use }0!=1) \\\\&= \dfrac{3!}{3!} \\\\&= \dfrac{\cancel{3!}}{\cancel{3!}} \\\\&= 1 .\end{align*} Hence $ _3C_3=1 .$