# On the probability of placing cups on saucers

A tea set consists of six cups and saucers with two cups and saucers is each of the three different colours. The cups are placed randomly on the saucers. What is the probability that no cups is on a saucer of the same colour?

What's wrong the the following solution ?

Update: The question is not a question any more since I managed to solve it in the right way thanks to @N. F. Taussig's answer, which helped me to spot few mistakes in the previous version. However, it can be seen as a solution based on an ordered sample space. See @N. F. Taussig's answer below for a solution based on an unordered sample space. See also here for a different answer to this problem. Here, I'm trying to solve it using the general Inclusion-Exclusion (just like @N. F. Taussig) principle.

Solution

My (lengthy) reasoning is as follows. Let the colours be azure, blue and cyan and let $$A_1$$ be the event "the first azure cup is on an azure saucer". Similarly, let $$B_1$$ = "the first blue cup is on a blue saucer" and $$C_1$$ = "the first cyan cup is on a cyan saucer".

Then the required probability is:

$$\Pr((A_1\cup A_2\cup B_1\cup B_2\cup C_1\cup C_2)^c) = 1-\Pr(A_1\cup A_2\cup B_1\cup B_2\cup C_1\cup C_2)$$.

Now there are $$720=6!$$ equally probable ways of putting six cups on the six saucers. In $$2\cdot 5!$$ of them $$A_1$$ occurs; in $$2\cdot 5!$$ of them $$A_2$$ occurs; and in $$2\cdot 4!$$ of them, $$A_1\cap A_2$$ occurs. The same holds for $$B_1, B_2$$ and $$C_1,C_2$$. We also have that $$\Pr(A_i\cup B_j) = \Pr(A_i\cup C_j) = \Pr(B_i\cup C_j) = \frac{2\cdot 2\cdot 4!}{6!}$$, for all $$i,j\in\{1,2\}$$. With the sum of probabilities of all possible double events being $$\frac{54\cdot 4!}{6!}$$.

There are 20 triples of events $$A_1\cap A_2\cap B_1,A_1\cap A_2\cap B_2,...$$. Of these, 12 events are made by two of the same colour and one of a different colour, e.g. $$A_1\cap A_2\cap B_1$$ and 8 events are all of different colours, e.g. $$A_1\cap B_1\cap C_1$$. For a triple with two colours we have $$\Pr(A_1\cap A_2\cap B_1) = \frac{2\cdot 3!}{6!}$$ and for a triple of three different colours we have $$\Pr(A_1\cap B_1 \cap C_1) = \frac{2\cdot 2\cdot 2\cdot 3!}{6!}$$.With the sum of probabilities of all possible triple events being $$12\frac{2\cdot 3!}{6!}+8\frac{8\cdot 3!}{6!} = \frac{112\cdot 3!}{6!}.$$

There are 15 events made by the intersection of four (simple) events, e.g. $$A_1\cap A_2\cap B_1\cap B_2,\ldots$$. Of these, 3 are double-coloured, e.g. $$A_1\cap A_2\cap B_1\cap B_2$$ and 12 are triple-coloured, e.g. $$A_1\cap A_2\cap B_1\cap C_1$$. For a double-coloured we have $$\Pr(A_1\cap A_2\cap B_1\cap B_2)=\frac{2\cdot 2 \cdot 2!}{6!}$$ and for a triple-coloured we have $$\Pr(A_1\cap A_2\cap B_1\cap C_1) = \frac{2\cdot 2\cdot 2 \cdot 2!}{6!}$$. The sum of all events made by the intersection of four events is $$3\frac{4\cdot 2!}{6!}+12\frac{8\cdot 2!}{6!} = \frac{108\cdot 2!}{6!}.$$

There are 6 events made by the intersection of five simple events, e.g. $$A_1\cap A_2\cap B_1\cap B_2\cap C_1,\ldots$$ and sum of probabilities of these events is

$$6\frac{2\cdot 2\cdot 2}{6!}.$$

Finally, there is also the last event to be considered $$\Pr(A_1\cap A_2\cap B_1\cap B_2 \cap C_1\cap C_2) = \frac{8}{6!}.$$

Putting everything together I found that

$$\Pr(A_1\cup A_2\cup B_1\cup B_2\cup C_1\cup C_2) = 6\frac{2\cdot 5!}{6!} -\frac{54\cdot 4!}{6!} + \frac{112\cdot 3!}{6!}-\frac{108\cdot 2!}{6!}+\frac{6\cdot 8}{6!}-\frac{8}{6!} = \frac{640}{720}.$$

Therefore, $$\Pr((A_1\cup A_2\cup B_1\cup B_2\cup C_1\cup C_2)^c) = 1-\frac{640}{720} = \frac{1}{9}.$$

Place the saucers on the table in the order: azure, azure, blue, blue, cyan, cyan.

Let's treat cups of the same color as indistinguishable.

There are $$\binom{6}{2}$$ ways to choose two of the six positions for the azure cups, $$\binom{4}{2}$$ ways to choose two of the remaining four positions for the blue cups, and $$\binom{2}{2}$$ ways to choose both of the remaining two positions for the cyan cups. Hence, the cups may be distributed without restriction in $$\binom{6}{2, 2, 2} = \binom{6}{2}\binom{4}{2}\binom{2}{2} = \binom{6}{2}\binom{4}{2}$$ distinguishable ways.

From these distributions, we must exclude those in which one or more cups is placed on a saucer of the same color.

Let $$A_i$$ be the event that an azure cup is placed on the $$i$$th azure saucer; let $$B_i$$ be the event that a blue cup is placed on the $$i$$th blue saucer; let $$C_i$$ be the event that a cyan cup is placed on the $$i$$th cyan saucer.

$$|A_1|$$: Since an azure cup is placed on the first azure saucer, there are two blue cups, two cyan cups, and one azure cup left to distribute to the remaining five saucers. They can be distributed in $$\binom{5}{1, 2, 2} = \binom{5}{1}\binom{4}{2}\binom{2}{2} = \binom{5}{1}\binom{4}{2}$$ distinguishable ways. By symmetry, $$|A_1| = |A_2| = |B_1| = |B_2| = |C_1| = |C_2|$$

$$|A_1 \cap A_2|$$: Since both azure cups have been placed on azure saucers, there are two blue and two cups left to distribute to the remaining four saucers. They can be distributed in $$\binom{4}{2, 2} = \binom{4}{2}\binom{2}{2} = \binom{4}{2}$$ ways. By symmetry, $$|A_1 \cap A_2| = |B_1 \cap B_2| = |C_1 \cap C_2|$$

$$|A_1 \cap B_1|$$: Since an azure cup has been placed on the first azure saucer and a blue cup has been placed on the first blue saucer, there are two cyan cups, one azure cup, and one blue cup left to distribute to the remaining four saucers. They can be distributed in $$\binom{4}{1, 1, 2} = \binom{4}{1}\binom{3}{1}\binom{2}{2} = \binom{4}{1}\binom{3}{1}$$ distinguishable ways. By symmetry, $$|A_1 \cap B_1| = |A_1 \cap B_2| = |A_1 \cap C_1| = |A_1 \cap C_2| = |A_2 \cap B_1| = |A_2 \cap B_2| = |A_2 \cap C_1| = |A_2 \cap C_2| = |B_1 \cap C_1| = |B_1 \cap C_2| = |B_2 \cap C_1| = |B_2 \cap C_2|$$

$$|A_1 \cap A_2 \cap B_1|$$: Since both azure cups have been placed on azure saucers and a blue cup has been placed on the first blue saucer, there are two cyan and one blue cup left to distribute to the remaining three saucers. They can be distributed in $$\binom{3}{1, 2} = \binom{3}{1}\binom{2}{2} = \binom{3}{1}$$ distinguishable ways. By symmetry, $$|A_1 \cap A_2 \cap B_1| = |A_1 \cap A_2 \cap B_2| = |A_1 \cap A_2 \cap C_1| = |A_1 \cap A_2 \cap C_2| = |A_1 \cap B_1 \cap B_2| = |A_2 \cap B_1 \cap B_2| = |A_1 \cap C_1 \cap C_2| = |A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1| = |B_1 \cap B_2 \cap C_2| = |B_1 \cap C_1 \cap C_2| = |B_2 \cap C_1 \cap C_2|$$

$$|A_1 \cap B_1 \cap C_1|$$: Since an azure cup has been placed on the first azure saucer, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have an azure cup, a blue cup, and a cyan cup to distribute to the remaining three saucers. They can be distributed in $$\binom{3}{1, 1, 1} = \binom{3}{1}\binom{2}{1}\binom{1}{1} = 3!$$ distinguishable ways. By symmetry, $$|A_1 \cap B_1 \cap C_1| = |A_1 \cap B_1 \cap C_2| = |A_1 \cap B_2 \cap C_1| = |A_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap C_1| = |A_2 \cap B_1 \cap C_2| = |A_2 \cap B_2 \cap C_1| = |A_2 \cap B_2 \cap C_2|$$

$$|A_1 \cap A_2 \cap B_1 \cap B_2|$$: Since both azure cups have been placed on azure saucers and both blue cups have been placed on blue saucers, there are two cyan cups to distribute to the remaining two saucers. They can be distributed in $$\binom{2}{2}$$ distinguishable ways. By symmetry, $$|A_1 \cap A_2 \cap B_1 \cap B_2| = |A_1 \cap A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1 \cap C_2|$$

$$|A_1 \cap A_2 \cap B_1 \cap C_1|$$: Since both azure cups have been placed on azure saucers, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have one blue cup and one cyan cup to distribute to the remaining two saucers. They can be distributed in $$\binom{2}{1, 1} = \binom{2}{1}\binom{1}{1} = 2!$$ distinguishable ways. By symmetry, $$|A_1 \cap A_2 \cap B_1 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap B_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1| = |A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap C_1 \cap C_2| = |A_2 \cap B_2 \cap C_1 \cap C_2|$$

$$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1|$$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and a cyan cup has been placed on the first azure saucer, the remaining cyan cup must be placed on the remaining saucer. There is one way to do this. By symmetry, $$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap A_2 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1 \cap C_2|$$

$$|A_1 \cap A_2 \cap B_1 \cup B_2 \cap C_1 \cap C_2|$$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and both cyan cups have been placed on cyan saucers, there are no cups left to distribute. There is one way to do this.

Thus, by the Inclusion-Exclusion Principle, the number of ways to distribute the cups so that no cup is on a saucer of the same color is $$\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1$$

Hence, the probability no cup is on a saucer of the same color is $$\frac{\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1}{\binom{6}{2}\binom{4}{2}}$$

• +1.Very neat! Going with unordered sample space is actually another viable option. Actually, you made me realise that I've made two mistakes: one in the calculation of the Pr. of intersection of 5 events and one on the intersection of 4 events. See the updated version. Commented Oct 15, 2018 at 7:30

What we do need here is a painting factory.

The two associated species equation are :

$$F = 1 + (R+G+B)(r+g+b)F$$ and

$$F = 1 + (Rr+Rg+Br+Bg+Gr+Gb)F$$

The generating function for the first "production line" is: $$f(x_R,x_G,x_B,x_r,x_g,x_b) = {1 \over 1-(x_R+x_G+x_B)(x_r+x_g+x_b)}$$

By Maple, the coefficient of $$x_R^2x_G^2x_B^2x_r^2x_g^2x_b^2$$ is $$8100$$ and it represents the number of all possible paintings of six china sets under the initial restrictions of color occurences.

The generating function for the second "production line" is: $$f(x_R,x_G,x_B,x_r,x_g,x_b) = {1 \over 1 - (x_Rx_r+x_Rx_g+x_Bx_r+x_Bx_g+x_Gx_r+x_Gx_b)}$$

By taking into account the second coefficient we obtain:

$$P= {900 \over 8100} = {1 \over 9}$$

• Thank you for your answer. What is "painting factory" ? Could you provide some details on it ? Commented Oct 18, 2018 at 7:02
• Suppose I have to generate two languages, one language $L$ of all strings $\{ Cc \}^n$ with $C \in \{ R, G, B \}, c \in \{ r,g,b \}$ and the second is L but avoiding pairs$\{ Rr, Gg, Bb \}$. I design two automatons that generate (or recognize) these languages. Then I associate generating functions. A "generating function" is some kind of abacus. For example, if I have two objects in a row $[a,a]$, I associate $x_a.x_a = x_a^2$. Commented Oct 18, 2018 at 11:30