Counting routes from home to office if you only return home if you realize you have forgotten something before you reach the office The following is how you can go to the office from home:
Home $\to$ Four Roads $\to$ Schools $\to$ Three Roads $\to$ University $\to$ Five Roads $\to$ Parks $\to$ Two Roads $\to$ Offices
You are forgetful minded. You may have forgotten something at home. You remember what you forgot at home, either at your school or university or in the park, and you go back to pick it up. Then the journey continues from the beginning again. You forget one thing at most during the day and when you reach the office you do not go back to take back what you have left. So how many different routes are possible for you?
I have tried this way:
Case 1: I remember what I forgot at school:
In this case, I can go to school in 4 ways and come back in $4$ ways and go back in $4$ ways. Then I can start the journey from there in $3 \cdot 5 \cdot 2$ ways.
Case 2: I remember what I forgot at university:
In this case, I can go to university in $4 \cdot 3$ ways and come back in $4 \cdot 3$ ways and go back in $4 \cdot 3$ ways. Then I can start the journey from there in $5 \cdot 2$ ways.
Case 3: I remember what I forgot at park:
In this case, I can go to park in $4 \cdot 3 \cdot 5$ ways and come back in $4 \cdot 3 \cdot 5$ ways and go back in $4 \cdot 3 \cdot 5$ ways. Then I can start the journey from there in $2$ ways.
Case 4: I go to office without remembering:
In this case, I can go to the office in $4 \cdot 3 \cdot 5 \cdot 2$ ways.
So, total ways would be $= (4 \cdot 3 + 3 \cdot 5 \cdot 2) + (4 \cdot 3 \cdot 3 + 5 \cdot 2) + (4 \cdot 3 \cdot 5 \cdot 3)+2 + 4 \cdot 3 \cdot 5 \cdot 3$ ways.
 A: Keep in mind that you add when two events are mutually exclusive, meaning that they cannot occur at the same time.  However, when choices can be made independently (such as choosing how to travel from home to school, school to home, then from home to school to university to park to office), you should be multiplying.
Case 1:  You have four ways of reaching the school.  You then discover that you have forgotten something at home.  For each of the four ways you could have reached the school, you have four ways of returning home.  For each of the $4 \cdot 4$ ways you could have gone to school and then returned home, you then have four ways of reaching the school, three ways of traveling from the school to the university, five ways to travel from the university to the park, and two ways to travel from the park to the office.  By the Multiplication Principle, there are
$$4 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$$
possible journeys if you realize that you have forgotten something once you reach the school.
Case 2:  You should have $4 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$ possible journeys (travel to university via school, travel home from university via school, travel to office via school, university, and park).
Case 3: You should have $4 \cdot 3 \cdot 5 \cdot 5 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$.  Do you see why?
Case 4:  You are correct that there are $4 \cdot 3 \cdot 5 \cdot 2$ possible journeys from home to the office via the school, university, and park if you do not realize that you have forgotten something until you reach the office, but you wrote the answer incorrectly when you added the four mutually exclusive cases.
Total:  Since the four cases are mutually exclusive and exhaustive, you obtain the answer by applying the Addition Principle to the four cases, which yields
$$4 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 +  4 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 + 4 \cdot 3 \cdot 5 \cdot 5 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 + 4 \cdot 3 \cdot 5 \cdot 2$$
