## Precalculus (6th Edition) Blitzer

The required solution is $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=-9499$
The combination formula is: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Applying this formulas into the provided expression, we get: $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=\frac{\frac{7!}{3!\left( 7-3 \right)!}}{\frac{5!}{4!\left( 5-4 \right)!}}-\frac{98!}{96!}$ $=\frac{7!}{3!\times 4!}\times \frac{4!\times 1!}{5!}-\frac{98!}{96!}$ $=\frac{7\times 6}{3\times 2}-\frac{98\times 97\times 96!}{96!}$ \begin{align} & =7-9506 \\ & =-9499 \\ \end{align} $\frac{{}_{7}{{C}_{3}}}{{}_{5}{{C}_{4}}}-\frac{98!}{96!}=-9499$