I guess there will be a standard answer to this question somewhere and wouldn't be surprised if this is a duplicate.
However, assuming you've got enough of each sort of candy (which 'big bag of' suggests), each child can have either of the three types of sweets. So, child one has 3 possible candies, child two has 3 possible candies and so on. For the first two children there are $3^2=9$ options this way (here you can still write down the options, if you like) and each successive child adds a factor $3$. Can you see why?
The result then would be $3^8$ ways to distribute candies (which equals $6561$, if I am to believe google). The questoin would be a lot harder if the amount of candies of each type would be restricted.
Your strategy is sound, but your answer is indeed incorrect.
Case 1: Sione receives two cookies.
Let $x_i$, $1 \leq i \leq 4$, be the number of cookies distributed to the $i$th friend. Let $x_4$ denote the number of cookies given to Sione. Since a total of ten cookies are distributed and Sione receives two,
\begin{align*}
x_1 + x_2 + x_3 + 2 & = 10\\
x_1 + x_2 + x_3 & = 8 \tag{1}
\end{align*}
Since each friend receives at least one cookie, equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of two addition signs in the seven spaces between successive ones in a row of eight ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, if we place an addition sign in the third and seventh boxes, we obtain
$$1 1 1 + 1 1 1 1 + 1$$
which corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 1$. The number of such solutions is the number of ways we can select two of the seven spaces in which to place an addition sign, which is
$$\binom{7}{2}$$
A particular solution of the equation
$$x_1 + x_2 + x_3 + \cdots + x_n = k$$
in the positive integers corresponds to the placement of $n - 1$ addition signs in the $k - 1$ spaces between successive ones in a row of $k$ ones. Therefore, the number of such solutions is
$$\binom{k - 1}{n - 1}$$
since we must choose which $n - 1$ of those $k - 1$ spaces will receive an addition sign.
Case 2: Sione receives four cookies.
Let the variables be assigned as above. Since Sione receives four cookies,
\begin{align*}
x_1 + x_2 + x_3 + 4 & = 10\\
x_1 + x_2 + x_3 & = 6 \tag{2}
\end{align*}
Since each friend receives at least one cookie, equation 2 is an equation in the positive integers with
$$\binom{6 - 1}{3 - 1} = \binom{5}{2}$$
solutions.
Case 3: Sione receives six cookies.
Let the variables be assigned as above. Since Sione receives six cookies,
\begin{align*}
x_1 + x_2 + x_3 + 6 & = 10\\
x_1 + x_2 + x_3 & = 4 \tag{3}
\end{align*}
Since each friend receives at least one cookie, equation 3 is an equation in the positive integers with
$$\binom{4 - 1}{3 - 1} = \binom{3}{2}$$
solutions.
Total: Since these cases are mutually exclusive and exhaustive, you can distribute the cookies in
$$\binom{7}{2} + \binom{5}{2} + \binom{3}{2} = 21 + 10 + 3 = 34$$
ways.
Best Answer
There are two different partitions for the cookies: $2, 2, 1$ and $3, 1, 1$.
The partition $2, 2, 1$ can be distributed in three ways (each child can the one with 1 cookie). The candies must all be given to this child, so we have a total of three divisions.
The partition $3, 1, 1$ can be distributed in three ways (each child can be the one with 3 cookies). For each of these three, there are four ways of distributing the candies to the other two children: each of the two children can be the child to receive more, and the split can be either $4,1$ or $3,2$. As such, there are a total of twelve divisions for this cookie partition.
In total, there are 15 ways we can divide the cookies and candies