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?

  • 1
    $\begingroup$ You chose a set of three positions for the blue, red, and green flowers. However, you have not multiplied by the number of ways that both the blue and red flowers are to the left or both to the right of the green flower. For each such choice, there are two ways both the blue flower and red flower can be to the left of the green flower (depending on whether the blue flower or red flower is in the leftmost of the three chosen positions) and two ways both the blue flower and the red flower can be to the right of the green flower. Therefore, you need to multiply your answer by four. $\endgroup$ Aug 29, 2019 at 21:56

1 Answer 1


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 $4/6$.

  • $\begingroup$ The problem consists of 2 cases, but I have included only 1 case for Rosie's garden $\endgroup$
    – user694067
    Aug 29, 2019 at 21:42
  • $\begingroup$ But why we don't we want the event that the flower is in the middle of the other 2? For example we have 6 places, and the green one is on the the second place, then the blue and red one are either on 1st and 3rd or 3rd and 1st, correspondignly $\endgroup$
    – user694067
    Aug 29, 2019 at 21:48
  • 1
    $\begingroup$ @AnnaKirilova If the green flower is between the blue flower and the red flower, then the blue flower and the red flower cannot both be on the same side of the green flower. Either the blue flower and red flower are both to the left or both to the right of the green flower. $\endgroup$ Aug 29, 2019 at 21:50
  • $\begingroup$ I'll explain more why to consider only the relative position of red, blue and green. $\endgroup$
    – Daniel
    Aug 29, 2019 at 21:52
  • $\begingroup$ @N.F.Taussig Ohh, I get it now, thanks for clarifying $\endgroup$
    – user694067
    Aug 29, 2019 at 21:53

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