probability that the blue and the red flowers are both on the same side of the green flower. Two sisters, Rosie and Daisy, bought two packs of flowers in their gardens. Each pack contains six flowers of different colours and the colours are blue,red,yellow,green,pink and white.
Rosie has a bigger garden and she plants all her six flowers in one row,
For Rosie's Garden, compute the probability that the blue and the red flowers are both on the same side of the green flower.
My attempt is :
$\frac{{6 \choose 3}3!}{6!}$ 
Where ${6 \choose 3}$ are the possible ways of choosing the $3$ flowers red,blue,green out of $6$  and $3!$ are the possible permutations of the 3 other flowers. And $6!$ are all possible outcomes.
I don't think this is the complete answer, could someone please identify what I have to add and explain why?
 A: Edit after rephrasing of the question. We have a row of $6$ flowers.
You want the probability that both the red and the blue flowers are on the same side of the green. One way to think of it is that all that matters is the relative position of these three flowers. There are $3!$ ways to order only the three of them, and in $2$ of these the green flower is in the middle, which is not the event we want. Thus, the probability is $4/6 = 2/3$.
A more detailed explanation
Suppose that you are considering all orderings of $6$ flowers. In your reasoning in the question you first pick places for red, blue and green and then permute the flowers, but not in a correct way.
We do something similar. Abbreviate each color by its initial. Suppose you chose positions $2, 3$ and $5$ for RBG, and positions $1$, $4$ and $6$ for YWP. One possible ordering in this case is
$$
 Y, R, B, W, G, P
$$
Consider now that positions of $Y$, $W$ and $P$ are fixed and we are allowed to permute $R$, $B$ and $G$. Then, the possible colorings are
$$
1. \quad Y, R, B, W, G, P\\
2. \quad Y, R, G, W, B, P\\
3. \quad Y, B, R, W, G, P\\
4. \quad Y, B, G, W, R, P\\
5. \quad Y, G, B, W, R, P\\
6. \quad Y, G, R, W, B, P
$$
and from these $6$ orderings, only $4$ of them obey the rule of red and blue on the same side of green (or how I rephrased it, 'green not in the middle'). Finally, notice that we obtain all possible orderings by summing over all possible positions and fixing some order for $Y$, $W$ and $P$. Since for each choice of position and order of $WYP$ we have $4/6$ cases of 'green not in the middle', then we conclude that the probability of this event is $4/6$.
