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
       $1 \le i \le 4$ 


The answer you got for the first question is right.
For the second, call a distribution bad if one or more of the $x_i$ is $\ge 7$. Our strategy is to count the number of bads, and subtract from the answer of a).
One can have $2$ of the $x_i$ equal to $7$. This can be done in $\binom{4}{2}$ ways.
Now we count the number of bads in which only one of the $x_i$ is $\ge 7$. Which one it is can be chosen in $\binom{4}{1}$ ways. Suppose that $x_1\ge 7$. Give $7$ candies to kid 1. The remaining $7$ have to be split between the four people, with none of 2, 3, or 4 getting $7$. There are $\binom{10}{3}-3$ ways to do this. 
We get a total of $\binom{4}{1}\left[\binom{10}{3}-3\right]+\binom{4}{2}$ bads. 
Alternately, we use Inclusion/Exclusion more explicitly. Choose one of the $4$ to give at least $7$ to, and give her $7$. We can distribute the remaining $7$ among the $4$ people in $\binom{10}{3}$ ways. But this double-counts the $\binom{4}{2}$ ways to give $7$ to two of the variables. So the number of bads is $\binom{4}{1}\binom{10}{3}-\binom{4}{2}$. 
