Precalculus (6th Edition) Blitzer

$-(\dfrac{21}{2})x^6$ or, $-10.5x^6$
Calculate the fourth term for $(x-\dfrac{1}{2})^{9}$ by using General formula such as:$(m+n)^r=\displaystyle \binom{r}{k}m^{r-k}n^k$ and $\displaystyle \binom{r}{k}=\dfrac{r!}{k!(r-k)!}$ This implies,$(x-\dfrac{1}{2})^{9}=\displaystyle \binom{9}{3}(x)^{9-3}(-\dfrac{1}{2})^3$ or, $=\dfrac{9!}{3!(9-3)!}x^{6}(-\dfrac{1}{2})^3$ or, $=(-\dfrac{1}{8})[\dfrac{ 9 \times 8 \times 7 \times 6!}{(3 \times 2 \times 1)6!}]x^{6}$ or, $=-(\dfrac{21}{2})x^6$ or, $=-10.5x^6$