Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$ 
Let $x_1, x_2, x_3, x_4$ be  nonnegative  integers.
(a) Find the number of solutions to the following equation: 
              $$ x_1 + x_2 + x_3 + x_4 = 14 $$

I got $17 \choose 3$ for this. Is that correct?

(b) Find the number of solutions if we add the restriction that $x_i \le 6$ for
       $1 \le i \le 4$ 


Use generating functions. The ways to pick each $x_i$ is represented by:
$$
1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = \frac{1 - z^7}{1 - z}
$$
Now you want four of those, and have them add up to $n$:
$\begin{align}
[z^n] \frac{(1 - z^7)^4}{(1 - z)^4}
  &= [z^n] (1 - 4 z^7 + 6 z^{14} - 4 z^{21} + z^{28})
             \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} z^k \\
  &= [z^n] (1 - 4 z^7 + 6 z^{14} - 4 z^{21} + z^{28})
              \cdot \sum_{k \ge 0} \binom{k + 4 - 1}{4 - 1} z^k \\
  &= \binom{n + 3}{3}
       - 4 \binom{n - 4}{3}
       + 6 \binom{n - 11}{3}
       - 4 \binom{n - 18}{3}
       + \binom{n - 25}{3} \\
\end{align}$
Here you have to be careful, the terms with negative upper index aren't present (no such terms are present in the sum, the first factor essentially picks out some of them).
In the particular case $n = 14$:
$$
\binom{14 + 3}{3} - 4 \binom{14 - 4}{3} + 6 \binom{14 - 11}{3}
  = 206
$$
