departamento de funcionários sábio e contagem de funcionários mais de 5

SELECT department_ID, count(employee_id) as '# of Employees' FROM EMPLOYEES 
WHERE DEPARTMENT_ID IN
 (
   SELECT DEPARTMENT_ID
    FROM EMPLOYEES
   GROUP BY DEPARTMENT_ID
   HAVING COUNT(*)>5 
 )
AND HIRE_DATE NOT LIKE '%JAN%'
group by department_ID;

  Department_Id | # of Employees
        1             7
        2             6
        4             9