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