I am stuck with the following math question and I am not sure the best way to approach it. As of now I have listen out all of the potential combinations have multiplied those by 5. I assume there is a smarter way to do this!
The question is such as: A professor is preparing an exam with five questions. Each question must yield a point ranging between 1-4. 1 point being the minimum and 4 being the maximum. The sum of all of the five questions must be 14. In how many ways can she subdivided the points for all of the five questions?
Let's work off the maximum of $4$. Specifically, we'll start by counting $4's$. We note that we can't have more than three.
Case I. $3\;4's$. Then the other two add to $2$, so the pattern is $\{4,4,4,1,1\}$. There are $\binom 53$ ways to place the $4's$ so $\fbox {10}$
Case II. $2\; 4's$. Now the other three must add to $6$. The patterns are $\{4,4,3,2,1\},\{4,4,2,2,2\}$ and we get $\binom 52\times 6=\fbox {60}$ and $\binom 52\times 1= \fbox {10}$
Case III. $1\; 4$. Now the other four must add to $10$. the patterns are $\{4,3,3,3,1\},\{4,3,3,2,2\}$ and we get $\binom 53\times 2=\fbox {20}$ and $\binom 52\times \binom 32 =\fbox {30}$
Case IV. $0\;4's$. As $3\times 5=15$ we can only have the pattern $\{3,3,3,3,2\}$ so $\fbox 5$.
In the end we get $$10+60+10+20+30+5=\fbox {135}$$
Note: While this kind of enumeration isn't really difficult, it is certainly prone to error so I advise checking.