Answer:
[tex]\large\boxed{\dfrac{7}{2}}[/tex]
Step-by-step explanation:
[tex]n!=1\cdot2\cdot3\cdot...\cdot n\\\\3!=1\cdot2\cdot3=6\\0!=1\to\text{from de}\text{finition}\\2!=1\cdot2=2\\1!=1\\\\\dfrac{3!+0!}{2!\cdot1!}=\dfrac{6+1}{(2)(1)}=\dfrac{7}{2}[/tex]
