Answer
The required solution is $\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=-2062$
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:
$\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=\frac{\frac{10!}{3!\left( 10-3 \right)!}}{\frac{6!}{4!\left( 6-4 \right)!}}-\frac{46!}{44!}$
$=\frac{10!}{3!\times 7!}\times \frac{4!\times 2!}{6!}-\frac{46!}{44!}$
$=\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2}\times \frac{4!\times 2\times 1}{6\times 5\times 4!}-\frac{46\times 45\times 44!}{44!}$
$\begin{align}
& =8-2070 \\
& =2062 \\
\end{align}$
$\frac{{}_{10}{{C}_{3}}}{{}_{6}{{C}_{4}}}-\frac{46!}{44!}=-2062$
