1. List the department IDs for departments that do not contain the job ID ST_CLERK, using SET
   operators.

	SELECT department_id
	FROM departments
	MINUS
	SELECT department_id
	FROM employees
	WHERE job_id = ’ST_CLERK’;
2. Display the country ID and the name of the countries that have no departments located in
   them, using SET operators.
	
	SELECT country_id,country_name
	FROM countries
	MINUS
	SELECT l.country_id,c.country_name
	FROM locations l, countries c
	WHERE l.country_id = c.country_id;
3. Produce a list of jobs for departments 10, 50, and 20, in that order. Display job ID and
   department ID, using SET operators.
   
	COLUMN dummy NOPRINT
	SELECT job_id, department_id, ’x’ dummy
	FROM employees
	WHERE department_id = 10
	UNION
	SELECT job_id, department_id, ’y’
	FROM employees
	WHERE department_id = 50
	UNION
	SELECT job_id, department_id, ’z’
	FROM employees
	WHERE department_id = 20
	ORDER BY 3;
	COLUMN dummy PRINT

4. List the employee IDs and job IDs of those employees who currently have the job title that they
   held before beginning their tenure with the company.

	SELECT employee_id,job_id
	FROM employees
	INTERSECT
	SELECT employee_id,job_id
	FROM job_history;

5. Write a compound query that lists the following:
   • Last names and department ID of all the employees from the EMPLOYEES table,
   regardless of whether or not they belong to any department
   • Department ID and department name of all the departments from the DEPARTMENTS
   table, regardless of whether or not they have employees working in them

	SELECT last_name,department_id,TO_CHAR(null)
	FROM employees
	UNION
	SELECT TO_CHAR(null),department_id,department_name
	FROM departments;