Answer:
[tex]\large\boxed{_4P_3>_4C_3}[/tex]
Step-by-step explanation:
[tex]\left\begin{array}{ccc}_nP_r=\dfrac{n!}{(n-r)!}\\\\_nC_r=\dfrac{n!}{r!(n-r)!}\end{array}\right\}\Rightarrow\dfrac{n!}{(n-r)!}>\dfrac{n^!}{r!(n-r)!}\Rightarrow_nP_r>_nC_r\\\\\text{because}\\-\text{the nominators are the same}\\-\text{the denominator}\ (n-r)!\ \text{is graeter than the denominator}\ r!(n-r)![/tex]
[tex]\text{Let's check our example:}\\\\n!=1\cdot2\cdot3\cdot4\cdot...\cdot n\\\\_4P_3=\dfrac{4!}{(4-3)!}=\dfrac{1\cdot2\cdot3\cdot4}{1!}=\dfrac{24}{1}=24\\\\_4C_3=\dfrac{4!}{3!(4-3)!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot1!}=\dfrac{24}{6}=4\\\\24>4\Rightarrow_4P_3>_4C_3[/tex]
