Answer:
50
Explanation:
Suppose that the kids get $x_1,x_2,x_3,x_4,x_4$ pieces of candy, respectively (each twin received $x_4$ pieces). Then we seek all quadruples $(x_1,x_2,x_3,x_4)$ satisfying
\[x_1+x_2+x_3+2x_4 = 6.\]We proceed by casework, based on the number of candies the twins receive.
If $x_4 = 0$, then there are 6 candies to distribute to 3 kids. This is like arranging 6 C's and 2 |'s (dividers), so there are $\binom{8}{2} = 28$ possible distributions.
If $x_4 = 1$, there are 4 candies remaining for the other 3 kids, so there are $\binom{6}{2} = 15$ possible distributions.
If $x_4 = 2$, there are 2 candies remaining, so there are $\binom{4}{2} = 6$ possible distributions.
If $x_4 = 3$, there are 0 candies remaining, for 1 possible distribution.
Adding all of these results from the separate cases, there are a total of
\[28 + 15 + 6 + 1 = \boxed{50}\].