Chapter 11 - Further Topics in Algebra - 11.6 Basics of Counting Theory - 11.6 Exercises - Page 1061: 23

$1$

$C(n,r)=\displaystyle \frac{n!}{(n-r)!r!}$. Factorials: $n!=n(n-1)!=n(n-1)(n-2)!=...$ $n!=n(n-1)(n-2)\cdots(3)(2)(1)$ $ 0!=1,\qquad$ by definition. --- $C(6,0)=\displaystyle \frac{6!}{(6-0)!0!}=\frac{(6!)}{(6!)\cdot 1}$ = $1$