Write SQL query to solve given problem: For the 3-credit course with the easiest difficulty, how many students get an "A" in that course?. Keep the solution inside sql tag ```sql [SQL-Query] ```
SELECT COUNT(T1.student_id) FROM registration AS T1 INNER JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T1.grade = 'A' AND T2.credit = '3' AND T2.diff = 1