Answer
$\dfrac{_7C_4}{_9C_4}=\dfrac{5}{18}$
Work Step by Step
Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ then the value of the given expression, $
\dfrac{_7C_4}{_9C_4}
,$ is
\begin{align*}\require{cancel}
&
\dfrac{\dfrac{7!}{4!\text{ }(7-4)!}}{\dfrac{9!}{4!\text{ }(9-4)!}}
\\\\&=
\dfrac{\dfrac{7!}{4!\text{ }3!}}{\dfrac{9!}{4!\text{ }5!}}
\\\\&=
\dfrac{7!}{4!\text{ }3!}\div\dfrac{9!}{4!\text{ }5!}
\\\\&=
\dfrac{7!}{4!\text{ }3!}\cdot\dfrac{4!\text{ }5!}{9!}
\\\\&=
\dfrac{7!}{\cancel{4!}\text{ }3!}\cdot\dfrac{\cancel{4!}\text{ }5!}{9!}
\\\\&=
\dfrac{7!}{3!}\cdot\dfrac{5(4)(3!)}{9!}
\\\\&=
\dfrac{7!}{\cancel{3!}}\cdot\dfrac{5(4)(\cancel{3!})}{9!}
\\\\&=
\dfrac{7!}{1}\cdot\dfrac{5(4)}{9(8)(7!)}
\\\\&=
\dfrac{\cancel{7!}}{1}\cdot\dfrac{5(4)}{9(8)(\cancel{7!})}
\\\\&=
\dfrac{1}{1}\cdot\dfrac{5(4)}{9(8)}
\\\\&=
\dfrac{5(\cancel4)}{9(\cancel8^2)}
\\\\&=
\dfrac{5}{9(2)}
\\\\&=
\dfrac{5}{18}
.\end{align*}
Hence, the given expression simplifies to $
\dfrac{_7C_4}{_9C_4}=\dfrac{5}{18}
.$
