In how many ways can points be assigned to five questions with one to four points per question so that the total is $14$? A professor must write an exam with $5$ questions. Question number i should give $p_i \in \mathbb{Z}$ points. The sum of the points must be $14$ and each question must give at least $1$ point and a maximum of $4$ points. In how many ways can he distribute the points on the questions?
I first turned $p_i= x_i - 1$ so I can get $0 \le x_i \le 3$ instead. Then I put the variable $x_i$ in the equation and got $x_1 + x_2 + x_3 + x_4 + x_5 = 9$. The total amount of combination is $C(tot) = C(5+9-1, 9)= C(13, 4)$. What I then did was find how many combinations there are when one $x$ can be greater than $3$, and then when $2$ can be greater than $3$. $C_1 = C(9,4)C(5,1)$ and $C_2 = C(5,4)C(5,2)$. So the answer is $C(13,4) - C(9,4)C(5,1) - C(5,4) C(5,2)$
What my answer sheet says is, $C = C(13,4) - C(9,4)C(5,1) + C(5,4) C(5,2) = 715-630+50= 135$ What I answered was $715 - 105 - 50 = 560$ How is $C(9,4)C(5,1)= 630$? Besides why did they do $+50$ instead of $-50$??
Can anyone help me ?

Our generating function is
$\begin {align*}
f(x)&=\left ( x+x^2+x^3+x^4 \right )^5\\&=x^5\left ( \frac{1-x^4}{1-x} \right )^5
\end{align*}$
So the answer is
$\begin {align*}
[x^{14}]f(x)&=[x^9]\left ( 1-5x^4+10x^8-h(x) \right )\sum_{k=0}^{\infty }\binom{-5}{k}(-x)^k\\&=\left ( [x^9]-5[x^5]+10[x] \right )\binom{k+4}{4}\\&=\binom{13}{4}-5\binom{9}{4}+10\binom{5}{4}\\&=\boldsymbol {135}
\end{align*}$
